 Hi friends, are you able to hear me? This is Tushar and just confirm whether my voice is reaching you all Yeah, so we will be starting at 6 30 just in case if you're having any trouble just let me know and Let me first start with sharing the screen. So just wanted to understand whether all of you are able to see this screen as well just confirm Are you able to see the screen? Yeah, so I Hope you're able to see the screen all of you Just confirm good so We will just wait for a couple of minutes and then we'll start And as you know the ground you have to put your mic always on mute because otherwise it interferes and there is a lot of noise So make sure that during Every minute of the session your mute mics are always on mute Whenever you require anything any doubt or something you can just unmute and Connect with us, you know all other fact factor numbers of Sentiment Academy would also be there it parallely so hence, you know Make sure that you know if you have any trouble you can As well uncle sir is also there so if I am busy in the session you can always reach out to him for any technical help Uncle are you there? So uncle sir will be here in some time and there I am responding over chat has been Okay, great. So all of you know the dates sheets are out And this time it is happening a little early So you know your dates, right? When the maths exam is scheduled so maths is on 12th of March but for R&R people you'll have a set of dates, I believe so Never mind. So Let's begin our this thing. This will be a two-hour session and from today onwards and Till 4th of Feb or till whatever, you know, when you will be writing your exams till then you will be available for any assistance possible so please make sure that you go through these sessions on a regular basis and We will be providing you with lots of resource material as well to Supplement your preparations. So if you are really You've already faced one set of exams in your schools If you are consistent with our sessions and you know the practice sessions and the mohawks and all that I Think are getting a hundred in both maths and science would not be much of a task. Yeah So, please make sure that you attend it consistently now the structure of this Revision program you already know that we will be revising the concepts first of all then the the focus would be on the type of questions which are asked in the board exams and We will align our approach towards them now and The second thing is you must also know lots of other resources available, which you must be aware of In what our model answers CBC has published all On their website, but we will also take you through in our respective sessions on You know, there will be discussions on previous year papers previous year questions Then we'll be discussing on and how people have You know such answers so and what are the possibilities of you losing marks? So we will be focusing on all of these parameters now if you have any discussion any doubt Let's use the session then you can always You know reach out to us But these sessions since you know that every session is of two-hour length So we try to make it as fast as possible because anyways, it's awesome It's a revision so we are not going into deeper details of each concept of it But yes as far as possible, we will be Trying to cover, you know as much details as possible should the case be that you're still Left with some kind of con some concepts which have not been touched or you really want some more Elaboration on that so you can always Come back to Okay, so let's begin. I hope Everyone is there so we will appreciate that if you join on time because again as I told you It's a two-hour session and we try to you know summarize everything so in those two hours so let's Today we are going to take up triangles and before that the regular CBC people, you know that the pattern has changed and This is how your question paper looks like So this is your standard question paper now standard mathematics. So if you see there are 14 questions My take on this kind of a paper is you know, uh, I actually it has become much easier now to attempt mathematics Because 20 marks are you know, uh, mcq. So if you see there are 20 mc, sorry 20 objective one mark each question Where in uh, you know, if you see there are 9 10 10 questions are mcq based So, you know, 10 questions are mcq based and then there are some fill in the blanks five fill in the blanks and there are five You know short answer a very very short answer type questions. So there are 20 questions I think within 20 minutes or at max within 25 minutes you'll be able to solve these 20 questions Then there are, uh, you know from 24 to 26. Yeah, so this is six question of two marks each So again, um, you know, it should take another 30 minutes. Let's say maximum to cover this Then there are eight three markers. So from 27th to I think 34. Yeah, so These are three marker questions. So again, you know in our R should be good enough for all of them, you know in 45 minutes in fact And then, uh, you know 45 minutes is good enough for these and then the last six again, which are four markers So let's say if you even if you spend one hour on each one hour on these So, uh, within two hours and 30 minutes you can actually wrap up the paper With some 30 minutes left for revision. So this is something which you need to do You need to have a strategy for this paper as well. So it's I am telling you again and again scoring Anything above 95 and if not 100 is very easy in this paper The you have to just, uh, you know, um, keep in mind the Structure so structure is very very very very very familiar and very very easy actually So hence, you know, you know, what kind of questions will be asked in fact By the time we'll be writing the board will be able to predict also. Yes the values there It will be slightly different, but it's quite predictable paper So hence the idea should be to finish the paper by two hours and 30 minutes Or at max two hours and 40 minutes and then the rest 20 minutes you can um spend on Revising something which normally people do not do and they lose marks on silly and calculation errors So I hope the structure is clear. So hence there are 21 marker Out of which 10 r mcq's five are Fill in the blanks five are very short answer type Then uh, you have a Six two marker Then you have eight three marker and again six four marker in total the paper will be off 80 marks duration is three hour Okay, so, uh, let's begin with this, uh, you know, so triangles again would be having around Eight marks question for sure if not more but then around eight marks One, uh, you know, so if you can see the sample paper itself, there will be one question on Let's say one marker Yep, so in this one marker, uh, let me see if there is one no, so in in mcq there is no But then in fill in the blanks, let's see if there is uh, yeah So if you see uh question number 13 one marker, yeah, it's there So you have to find out the ratio of the areas if you see this is very very important Every year they are asking question on If side ratios are given then find the ratio of the areas or the vice versa Right, so hence you cannot really miss out on these kind of questions So please be very very careful the only thing you will make mistakes will be during calculation So you need to just be a little bit more cautious Then here again, there is one marker. So, you know, so find the length of ab and uh Um, I think in the figure angle a cb cda a cb cda, yeah So an ad is three centimeters. We have to find out the length of ab So, you know, there are there will be at max two marks of one Sorry, two questions of one marker one mark each Then here is question number 23, which is again based on, you know, the triangle chapter. So two marks here So total four Let's go to section c section c may if you see uh here, uh Uh, yes, so bc is perpendicular. This is this you can treat as You know, so it is, you know, kind of mix of both, but let me see if exhaustive Uh triangle question is there only only triangle already. Yes. No, this is also So no, so till there no none and uh, this is construction. So this is also not there And uh, yeah, so 36 if you see this, you know, line drawn parallel to one side of the triangle to intersect The other two sides in distinct points Then the other two sides are divided in the same ratio You know, this is the famous Thales theorem this bpt. So they are directly asking the theorems prove So all the theorems become very very important. So you can see four plus two plus two. So eight marks from the Triangle, okay, so hence triangle will have at least eight marks out of 80. That means 10 percent of the Storal score. So you can't really, you know afford to miss out on this and triangles Are a little tricky. So never mind. So what we are going to do is we are going to run through the uh Summary of the entire chapter. Do not mind. Uh, sorry. Do not worry. This is uh, the entire video will be like, you know, recorded And uh, this ppt also would be shared with you. So hence you can Use it later. So first thing first, uh, they have defined similarity It could be a question on one more question in your board So two figures having the same shape but not necessarily the same size are called similar figures, you know already So they can give you a few choices. For example, if they are let's say Uh, which of the following are always similar? So they can give you two circles Let's say and they can give you uh, two Squires And let's say they are saying two uh, two hexagons hexagons and uh, hexagons and then d Two, uh, equilateral triangles equilateral Triangles so what triangles which one of them are not which one of them is not similar not necessarily similar So what would be the answer, right? So this could be on the very basic um So, uh, RO is not able to hear anything. So that means your speakers are not proper. Uh, uncle, please take care of that So, um, yep, uh, guys. So answer would be c correct. No, so Uh two circles Two circles two squares two hexagons. So, you know, two hexagons not need not be Need not be similar. So if they are two regular hexagons, then definitely but then it's not mentioned over there So please understand two figures having the same shape But not necessarily the same size are called similar figures. The application could be this I hope you understood this Next is the all congruent figures are similar, but the converse is not true Yes, you already know this it can be a true false question if at all they want to, you know, uh You know ask you like that. So all congruent figures are definitely similar But converse is not true. Why because uh, if you have two circles of different sizes, let's say this two different radius The other one doesn't look like a circle. Let me Yeah, yeah, so this is a circle now. So these two circles definitely are similar They are similar, but definitely they are not Congruent because the sizes are the same Okay, next So two polygons having the same number of sides are similar If yeah, so this is nothing but we are talking now talking about criteria of similarity Yep, so, you know Any two Geometric figures will be similar provided Point number one their corresponding angles are equal and point number two their corresponding sides are Proportional now it is true for any Geometric figure, right? So let's say if you have a pentagon And another like that So both pentagon are similar only when let's say a b c d and D and e and this is let's say p q r s and T right so they are similar only when Corresponding angles are equal so hence that is very very critical And here is where most of the errors happen. So a must be equal to p and Yeah, and b must be equal to q c must be equal to r D must be equal to s and e must be equal to t and the corresponding sides Must be proportional. What does it mean if you already know this? This is a b upon so a b upon a b upon p q is equal to b c upon q r and so on and so forth So finally e a upon t p like that. Okay, so hence If this is true, then these two are similar okay so Same holds for triangles but triangles one beautiful part is you don't need to you know Triangles the only geometric figure which is possible where you don't need to Have both together true And for triangles the best part is Either one or two should be true Right, that means either one or the two should hold and the triangles are similar It's possible only for triangles, you know not for any other geometric figure And this is because the triangle is the smallest possible polygon Yep, so this is So for triangles, we know we need to just prove either this or this But for all other figures, we need to prove and you know, we need to prove both of them to make Make sure that they are Similar, okay Now let's come to the theorem. Now. This is the question which is given in your The question the you know sample paper as well. So this is the famous Uh Thales theorem or basic proportionality theorem if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points Then the other two sides are divided in the same ratio. So hence be very very Thorough with the proof you cannot mess Out in this, okay, so hence this is the triangle and this is a line parallel So quick proof, you know how to do it a bc And this is let's say d and e. So how do we prove it? So basically what you do is you drop perpendicular from or you basically join these two points c e and Be right and then we use the concept of areas to prove This one. Yes. So what do we do? So we we we say we say area of triangle a d e a d e divided by area of triangle d e b d d e b area of triangle d e b Is both are is is nothing but a d upon d b why because uh Area with same altitude. So if you see this is a these two triangles are having same altitude and that altitude is nothing but if you see Let's say this one edge So hence you can direct you can say area of a d e divided by area of d e b triangle d e b is a d by d b Okay, and you have to mention the um mention the reason as well So uh many people skip the reasoning part and hence can miss out on the marks. So to write uh area of triangles or ratio of ratio of areas of triangles with Same height Same height Is equal to ratio of their bases This is the underlying philosophy here Right. So hence similarly you can also say um Yes, so hence uh, so and similarly you can say that area of triangle a d e divided by area of triangle d e c d triangle d e c Again will be equal to a e Upon e c same reason. So and then you mention this as one Mention this as two Okay, so that means if you divide one by two, so you say one divide by two Right, so what do we get and I'm not writing this all the steps But you have to play, you know make sure that you write everything here and here it is nothing but area of triangle d e c divided by area of triangle d e b Now this is equal to um, sorry, uh Oh, wait a minute. Uh, where did I yeah, so basically? Oh, okay. Okay. Wait a minute. So ad ad by d e ad by ad by d b is equal to oh, sorry You don't need to divide it actually. So what you can say is These two ratios are same. Why because if you see You can see area of area of triangle d e b Is equal to area of triangle d e c. Why is this? Again d e b and d e c are having same base Same base and same base is equal to d e and between same parallel between Same Parallel correct. So hence this is three Okay, so from one two and three clearly what happens LHS of both one and two becomes same, right? No so hence from one two and three from One two and three You get ad by d b Is equal to a e by e c Okay So this is what Is thaili's term is what you have to prove, right? So please remember, uh, how to go about it Okay, so this proof is Important, let me write it here So do not forget to revise it before the exam. Okay, and the application obviously We'll be there. Okay Uh, give you a minute. Okay, Veda. Give me a minute. What what do you want? Tell me If you want to uh, if you're writing it. No, don't worry. We anyways, I'll give you this entire ppt So you can you know take it as a note. Okay So in uh, what are important, uh, uh, things here would be please, you know, so make sure you draw neat diagrams So diagrams will be very very important Diagrams they should be neat Yeah, do not do any overwriting anything of that sort. Okay, and uh, you know label label them Label them properly So marking scheme says all this so label them Yeah, well labeled diagram neat answer attracts good marks. Okay. Okay. So let's move on. So this is uh, The converse of uh, bpt. Yep, so Basic proportionality theorem. So hence Thing is basic basic Proportionality proportionality And uh, you will see that they will not uh, usually Ask the names of the theorem as in, you know, they won't say okay prove basic proportionality theorem They will give the theorem And then ask you to prove it. So hence you don't really need to remember it But then yes, it's good practice to Remember and it is also called Thales theorem. We have already spoken about it Thales theorem. Okay Now if a line divides any two sides of a triangle in the same ratio Then the line is parallel to the third side of the triangle. This is converse of it We have done it multiple number of times and uh, what is important here is that we use a Concept called contradiction So this is the best part about proving so if you know the first proof the converse majority Number of times major number of times you can actually prove the Converse using contradiction. So hence what does it mean? So hence we have a triangle again and and this thing says that okay if Uh Line is there which is dividing the ratio the triangle a b c Into equal ratios That means it is given that a b It's given that a b by let's say. Oh not a b a let's say this is d and this is e so ad by db is equal to a e by ec it's given and you have to prove that prove that de de is parallel de Is parallel to bc bc This is what you have to this is the converse. They can also ask you to prove this Do we have to even learn the statement also? Uh, I don't think so. They will not ask you. Okay. What is basic proportionality theorem But it's always, you know, it's not very hard actually to remember it as well to answer your query with him. So it's always good to know the you know, uh Statement as well, but then you can frame it in your words and write if at all they are asking it. Okay Now so de is parallel to bc and whenever such things Let's say if you're not able to articulate articulate any statement or something you can always use expressions or let's say, you know Diagrams to explain and mathematically define it. It should be You know, it should serve the purpose Uh, isn't it mandatory to write the statement before starting to do proof? Uh, not necessarily the question is given. You don't need to copy the same thing again If you really want it, you can summarize like this. For example, what did I do here? I drew the triangle and explain what is given and what is to be proven So this is nothing but in a way I wrote the same statement but in mathematical language Okay, so this is what you have to start with the you know, the usual process of writing given so do not after some time when we'll discuss the um Everything will be done. Just just just hold on guys. So yes. Yes. It is to answer your query You can define like that. I will show you how people have done it So hence given you have to mention so all this have to be written and given then the next step is Uh, to prove or to show please write in that order itself And then uh, you know, it is advisable to draw the diagrams using pencils now you have enough time for all this So, uh, you can uh, you use that okay Now you use the draw diagrams using pencils and all and then to prove after to prove So, uh, you know, you can do construction construction construction and then finally The proof part. So these are the four parts like that you should you know, uh Let's write your answer. So now in this proof, you know, how to do it Contradiction proof, right? So hence, what do you do? You will assume that let's say a de is not parallel to pc So let's say de is Not parallel to bc Okay, so do not write these symbols because you know, so you let me just not use it only So you you should write let's say de is not parallel to bc Okay So let's say Then let then you can say let's say let's say De dash Is parallel to bc and hence you have to also draw de dash. So let's say you're drawing like that and mention e dash Okay, therefore by bpt, you know By bpt by bpt. Yeah Yeah, it's always, uh, you know, uh, uh, don't write bpt You should write basic proportionality theorem, but then you know, it also depends many times people are considered if you're writing bpt. Oh, sorry What happened? Yeah, oh just a minute guys, huh? So, um So let's say de uh dash is parallel to bc. That means from bpt Oh Yeah, so From By Bpt you have to say that ad upon db is equal to a e dash upon e dash c right, but but But you should write but ad by db is actually equal to a e by ec this was given Right, so then it's always advisable to you know reduce your writing time You write equation number so you can write now write from one into What do you get a e dash A e dash by e dash c is equal to a e by ec Okay, this is only possible. You can you know that this is only possible when when e dash and e coincide Okay, e dash and e coincide and hence our assumption that de is not parallel to Um bc is wrong and hence de is actually parallel to bc So hence i'm not writing those steps. You please make sure that i'm just trying to you know give you the The the quick revision part of it. Okay, so let's move on. There are 30 slides in this and there's a few questions also So the internal bisector of an angle of a triangle divides. This is also important term important. Let me write it here And they important they can ask you the proof of this many times. They've asked it and you must know how to prove it What does it say? So there is a triangle and There is a triangle. Sorry. There is a triangle here Okay, and now Um You are joining it or let's say you have a you have an angle bisector You have an angle bisector here Okay, so hence now, uh, this is X and this is X So a b c D. Okay, you have to prove what that ab ab by ac is equal to bd by dc right, so remember, uh The you know the statement and What exactly they mean so hence this by this so ab by bc ac ab by ac. Sorry is equal to bd by dc that's what they are asking to prove and you know, uh in this case There is you know, she's always good whenever you're writing, uh, or let's say doing a revision So you mention wherever let's say some construction. This is a problem where construction is required Okay, what is the construction? You know, you um for the positive space. I'll do it itself So you X you extend ba like that Okay, and then what you do and how do you extend it you make sure that a so hence construction here is Let me write construction here is After you write given to prove and all that then construction. What is construction? Uh, ba produced. So this is how you should write ba produced to e such that such that A e is equal to Or or rather, uh, no in this case, you should not do this in fact the construction you can this do this as well But the usual given proof is you have to extend or draw Yeah, like that you draw You draw. Sorry. So this construction this could be uh, this you could have done this way also But then your ncrt style. So what do you do is construct c e parallel to A b you construct this and then let this Join both both ways are possible. So let's join this Okay So c e parallel to ab drawn Oh drawn whichever way suits you you can do this. Okay, and what? Uh ba produced produced To meet meet C e at e this is how you should write A d not a visa c parallel to ad c e parallel to a oh, I'm sorry. Yeah, I'm sorry ad Yeah, be very careful because these things will lose marks. Very good. Yes, right then what? Then we already know that. Okay. If this is the this is the case you can denote in the Diagram as well use the diagram as much as possible. So, you know, for example, I'm writing x here y and then you state it here So hence you'll you'll say this is e. So hence angle Be C is equal to angle ba d Is equal to x Right. Why it is because corresponding corresponding angles Okay, and similarly angle d ac is equal to angle a c e Again, this is because this is equal to x and this is alternate interior Now I'm writing short form. You please do not write that there alternate interior angles Okay, that means what ac is equal to a e Isosceles triangle Right. So ac is equal to a e that is what we wanted to prove now hence What will happen that means? Yes So now in triangle b e c By b p t what will you write by b p t and you know, uh, ac is equal to a e So you deploy it here and hence you will get this implies ba by ac is equal to b d d by dc hence proved Okay, so This is the proof for this very this is an important proof. It might be there for a war marker Okay, next so converse So I hope so far so good. No one has any doubt in this Converse again is very very, you know, simple similar thing you have to do So what does first of all you must know how to write the statement So hence if a line through one vertex of a triangle divides the opposite side in the ratio of other two sides Then the line by sexy angle of the work angle at the vertex. So hence Let us first draw a triangle So this is a triangle and it is said that a b c a b c and There is a line ad Such that so what's given given part is a b upon ac Is equal to bd upon dc okay Is the question six in question very n crt uh Question six in this is no question six. These are all theorems which we are revising guys So, okay. Now. What happens here a b by ac is equal to bd by dc So, uh, you have to prove, huh? So hence what is to be proven to prove to prove. What do you need to prove? You need to prove that Uh a angle B ad is equal to angle c ad. That's what you have to prove Okay Okay, guys. Yep. So hence, how do you uh do this? again, you have to do a construction And what you need to do is you can again draw, you know parallel line Like that and extend so i'm not writing that i'm just explaining this one to you and you can write all of that Okay, so what is this now? So let's say this point is e again Sorry for the small space. Okay. So e again. So if they are parallel that means this uh Yeah, so hence, um Very good. So hence if you draw parallel then what happens? Let me write it here so clearly ab upon um Yeah, so they're parallel then by bpt by bpt we can say ab upon ae is equal to bd upon dc Right ab upon ae is equal to bd upon dc, but it's why why this because it's because of bpt, but we know but ab upon ac is equal to bd upon dc. It's given By one and two One and two what can you infer you can infer Ae is equal to ac Right a is equal to ac the moment a is equal to ac then you can write this This as x this as x Okay, so if these these are two are x then you know by corresponding angle this also will be x And by alternate interior angle, this also will be x. So hence our purpose is solved, right? So, but you have to write it in this fashion that A is equal to ac that means angle Angle A e c is equal to angle a c e Right also Angle a e c is equal to angle D a d and you have to write corresponding angles also angle a Sorry angle d a c is equal to angle a c e. What is this alternate interior Angle so again you write three four And five so always put, you know the equation number. So hence and then finally you can write from three four and five What can you say? angle b a d Is equal to angle d a c or c a d hence proved Any doubt guys in this Right, so you now know in angle by sector theorems You have to draw construct lines parallel to the You know the bisector line Okay, great. So let's move ahead now. This is not there in your this thing This is not there the external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle now, but If they give you as a problem If they give you as a problem, then you'll have to do it. What are they asking? They're asking this that if There is a triangle And you know there is an external external bisector of this So let's say I have this triangle okay, and This is the external angle. Let's say this is a This is b. This is c and the bisector of this is let's say meeting Yeah, this point is d Okay, this is the case now. They're asking that that um Given that this is x x and this is also x prove that a d by I'm sorry a b by a b a b a b by ac a b by ac is equal to bd by cd Okay Yep So hence guys, uh, they will not ask you directly, but they can give you as a problem to Solve. Okay. So hence that's you know, so you must be see you should not leave any stone unturned, you know, so do not Say, uh, think only okay because it's in in Um, syllabus or not in syllabus, but this can be asked as a question to you to prove So in this case, you need to remember that what do you need to do again? This will not be without a construction. So I'm I'm writing here This problem requires construction What is the construction now again? You have to drop parallel lines like that. Okay, so you have to drop parallel from c Right, let's say this point is e Okay, so the moment you do this what else do you get so this angle becomes x Okay, and this angle also becomes x Again by corresponding an alternate Right, so I'm not because there are lots of other things to cover. So I'll be very quick in solving this So if you see from here a e a e A e is equal to ac Why isosceles triangle opposite angles are equal so a e is equal to ec Uh, sorry a is equal to ac. Yep a e is equal to um A e is equal to ac. Yeah a is equal to ac now by bpt again, you can say b e Be upon or rather the corollary of bpt. So hence I know I I hope you know, you remember corollaries as well Corollaries of bpt is what so let's say if this is the triangle and You know, this is x by y and m by n. Let's say these are the ratio Then you know the corollaries are like this x plus y by y Is equal to m plus n by n. Yeah by component of dividend. Oh, we had done that. So but anyways, you remember this otherwise also x upon x plus y will also be equal to m upon m plus n All these things you must know And y upon x plus y will also be equal to n upon m plus n. So all these ratios are up bd by bc bd No bd by cd ab by ac How do you remember you go from one point? From b to d and then come to c like that bd by dc Okay, guess 57. What's your name? Why are you unknown? Anonymous, what's your name? Okay, uh, so please mention your name so that we can you know, uh Identify you. Okay. Anyways, so this is yeah, so hence what will happen here So I'm you know, we will be using these corollaries. I'm just Okay, let me let I'm putting it here itself Yeah, so now a e is equal to ac given so b b a b a upon e a so we can write this as b a upon e a b a upon b a upon medha oh b a upon e a is equal to bd upon cd Is that okay? Right by bpt and now e a is equal to ac so hence I can replace it by b a by ac is equal to bd by cd hence proved cd Okay, fair enough any doubt in this So this is the uh external bisector theorem, so this can be asked as a question moving ahead Now this you already know midpoint theorem very very easy No need to you know, so you know that if this is a triangle abc What does it say the line drawn from the midpoint of one side of a triangle is parallel of another side by six the third side So let's say point d is the midpoint And you draw a line parallel meets at e. You know e is also special case of bpt Uh, what is special case of bpt? Yeah, she is uh focus here now. So hence uh de So d is the midpoint. So hence by bpt. We know ad upon db will be equal to ae upon ec isn't it So d is the midpoint that means ad by db is one by one Is equal to ae by ec So hence we get ae is equal to ac Ha, especially case of bpt very good, right? So ae is equal to ec So this is what is midpoint theorem here. So hence midpoint is there and uh, you already know, which is not You know, you would have learned in ninth grade that de is also half of bc So keep that in mind You can use this if required at all Okay So, okay, let's move ahead the line joining the midpoints of two sides Is parallel to the third side. This is again converse of midpoint theorem No brainer again. So if there is a triangle abc abc And it says line joining the midpoint. So let's say d and e are the midpoint and you join this So hence they are midpoints given you have to prove that de is parallel to bc. This is this has to be proved Okay, so by converse of bpt you can say that ad by db anyways is equal to ae by ec right And this is equal to one by one So by converse converse of bpt now many prop many students face this issue of remembering which is the First theorem in which is the converse of it. Now bpt talks about bpt says that there is a parallel line already So the moment there's a parallel line, there will be two equal ratios. So remember like that So parallel line leads to ratio is bpt. So the converse is equal ratio leads to parallel line Right. So, you know, you have to have some trick of remembering. So many people get confused. Which one is the You know the the main theorem and what is the converse? So bpt is Parallel line is there. It cuts the two sides. It cuts the two sides in equal proportions. That's bpt Basic proportionality theorem proportional the word proportionality is there. So hence you will get proportions there So that's basic proportionality theorem, right? So by converse of bpt now already proportion is there. That means the line will be parallel, right? So hence we said de is parallel to bc clear, so this is Okay now Diagonals of a trapezium divide each other proportionally This can this can be given as a, you know, theorem to prove or some an application based on this. So let's say these are the sides a b c d a b and c, uh, sorry a c and bd are the diagonals Diagonals of a trapezium divide each other proportionally. That means what does it mean? It means you have to prove a o upon oc is equal to b o upon od now do not mess with the order Order is very important. You can't say a o by o a o by oc is equal to od by ob this will be wrong Okay, so a is on you know a on top then o then c then o then b like that everything just froze what froze Everything is fine. All of you are able to any technical issue guys. Please put in the chat box All is well Any any any any any issue Good. Okay. Very good. Kavi. Uh, just log out log in Any any technical issue, any way you feel just log out log in. Okay Chittage no comments. Okay now guys, uh, so how do we do again? This is important in terms of again construction Remember the construction here here also there will be a construction What is the construction you have to draw a line parallel to? Let's say this point is e Okay, now you can use bpt and then You can do Okay, so hence Uh, the moment you draw parallel lines. So hence, what do you get in triangle? a o sorry not a o a dc You will get a e upon Ed is equal to a o upon oc By bpt Okay And in the other triangle. So see uh, keep in keep in mind. So a e by ed is not required a o by oc is required And the other thing is b o by od So b o by od choose that triangle where b and od are there Okay, so and that triangle is d a db. So hence triangle a db If you see it's What do you what do we see a e upon same ratio you take a e by ed Is equal to ob by od Right same bpt and now if you check this is what we wanted both lhs are same. So rhs will have to be same Okay, so a o by oc is equal to b o by od and hence we get this result The converse is also true Right that and this is the next slide converse is if the diagonals of a quadrilateral. So let us have our quadrilateral first So these are oh there becomes a regular Try not to draw regular figures because otherwise you'll get confused. So let's say This is the uh Yeah, in the diagonals of a quadrilateral divide each other proportionately. So or proportionately. So let's say This is the case a b c d Oh Now this is d right. It's given that a o by oc is equal to ob by od then you have to prove that a b is parallel to dc And no brainer again. You have to just draw perpendicular. Sorry parallel line name it e Right. So hence it is parallel. So you're so parallel to one of them. So let's say i'm drawing construction What is the construction draw e o parallel to dc one parallel line? I can definitely do draw, right? Okay, now, uh Can't we use similarity in which question? uh in this Oh If yeah in there's no problem in using similarity You can use similarity as well. No problem. But then why kill a rat using a your bond So Let's use something which is very very simple. So b p t right e o is directly Sorry e o is parallel to e o is parallel to dc. We constructed Then this means what this means do do Do by ob will be equal to d e d e by e a d e by e a hey, don't do this. Yeah, who's using annotation? I'll throw you out of the class I told you do not distract Stay away from this I know who's doing it. I'm not trying to tell you publicly I'm arranging the class on my pc and I can't hear anything. Uh, you know check your issue. You know this thing. Um Speakers Arjun everyone else is able to do it so carrying on Do by ob Do by ob is d e by ac. That's what Sorry d e by e a and we already know from this that do by ob Is so I'm writing it here. So here is what I'm continuing here So do by ob from the given condition is o c upon a o Right from the given condition here And here you can see what do I know This means o c by o o c by a o is equal to d e by e a From the first two conditions o c by a o is equal to d e by e a and this by Converse again converse of b p t What can we infer we can say that ab is parallel to e o Okay, ab is parallel to e o but e o was parallel to dc from here. See e o was parallel to dc therefore By combining these two this one And this one you will get ab is parallel to dc Okay, so see guys what I'm trying to tell you is this please do not do these questions in this short form Since this is a you know revision class So I am just going through the theorems one by one when we'll do do the questions down there Then we'll write all the steps. This is just for your You know for you to refresh your you know knowledge in this Okay next Any line parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally again Direct fallout of b p t. So they're saying that there is a trapezium And you have a parallel line Let's say here and whenever such kind of problems are there as I told you never draw very regular Cases, so let's say, you know, I take take cases like this So that your you know mind is not confused. So ab cd And let's say de and f So this is parallel and you have to do what you have to say you have to prove that a So what is given? given is a E f is parallel to ab parallel to cd What you need to do is prove that prove that a e by e d a e by e d okay a e by e d is equal to b f By f c this is what you have to prove How to do it very simple again you have to again do some you know Again the the moment there is some ratio, you know what to do You have to do Bpd use Bpd. So hence what I do is I construct this line parallel Make a triangle. So you what one thing you need is triangle. So hence what I do is I draw construction construction will be a g parallel to b c Okay, I draw Do this what what does it become a g? Cb is a parallelogram It becomes a parallelogram Right, then clearly if you see Yeah, so And let's say e f g edge this point is edge So this point is edge Okay Now, what do we get? We get by b p t sorry by b p t by b p t we get a e upon e d is equal to Yeah, a e upon e d is equal to a h upon h g by b p t correct. Now if you see a h f b and H g c f Both are parallelogram Why because opposite sides are parallel Correct. That means a h is equal to b f Opposite sides of a parallelogram are equal And h g will be equal to c f Right, and that is what we wanted. So hence you can now say a e by e d is equal to b f by c f Right, this is what you wanted to prove and it's done Very good Draw a dia Oh, you can do that as well. No problem Draw dia. So what she is just saying is draw a diagonal Like that and then again you will be p t you can do that also no problem Meda said that Yeah Okay, Meda. Yes, you can do multiple ways. No problem. Whichever thing you think is easier for you. You can go ahead and do it Okay, uh Next is if there are more parallel lines are intersected. So this is called intercept theorem If three or more parallel lines are intersected by two transversals then the intercepts made by them on the transverse Proportional, what does it mean? So there are one two and three that's the minimum of three lines are there and there are two transversals Like that So let's say this is l. This is this is l. This is m. This is n and this is Let's say point a point b point c point d point e and point f So you have to prove given what is given l is parallel to m is parallel to n this is given this is Given you have to prove that ac upon ce is equal to bd upon df Okay, so how do again so you can do the transversal thing if you want this diagonal thing again here also So join this Let's say this is g Okay, so clearly what do we know? Uh, we know that A c by c e will be equal to b g by g e again by bpt Right all are parallel so hence by bpt we can say and similarly we can say b g by g e Is equal to b d by df And hence prove One and two if you combine them you will get the desired result Like that Okay, you can use the previous question as all the parts are to medium Yes, but then uh Yeah, you can do that no problem Okay, but then don't you know it's it's always better to do that never use that. Okay, since it is a trapezian man Uh, because cbc marking scheme may not like that. So hence do the you know way it has been instructed Okay, uh now okay now comes after the first part if you if you see the triangle chapters I didn't tell you before is you know, there are three basic elements one is bpt So there will be problems related to bpt always remember second, there is similarity of triangles and third Pythagoras theorem So these are three three major You know this thing In your triangles chapter. So questions would be Pythagoras theorem is a very last year It was asked to prove the Pythagoras theorem Yes equal intercept Not equal intercept intercept theorem. They're not equal Intercepts are not equal. They're proportional to each So, you know, the first part is bpt second is similarity and now so a similarity criterion So what is uh, uh a similarity criterion guys? So hence there are two triangles and uh And what the saying is if how what is a a similarity a similarity means now we are talking about the criteria enough So we learned that there are two criteria for any two You know figures to be similar their ratios their angles should be equal corresponding angles should be equal And the ratios of these sides should be proportional. So a b c def Right and in and I told you in case of triangles In case of triangles, we don't need both We just need to prove any one of them and it is done. So that's And you can prove it from here itself. Let's Just a minute a d yeah, this is d So a b c and def are similar and what is given that angle a is equal to angle d And angle b is equal to angle E and c is angle f. This is given Okay, so I am I have not seen You know people asking the proofs of the similarity criteria But why take chances? So hence, you know, uh, we will you know learn the proof as well as fast as possible So the first one is always remember the order also is very important a a a similarity has to be known first Once you know this you will use this particular similarity to prove other similarity So hence in this case, what do we do we consider? So, uh, how do we prove similarity? So we already know the angles corresponding angles are equal Let me just make it a little specific and this is equal to this And this is equal to this now. How do we prove the similarity? So if you somehow prove that the ratio of these sides are also equal then you are done So let's say, uh, this is a thing. Let's say I have case one Case one is let's say All the sides are you know a let's say, uh, we are saying a b is equal to Ed there are three possibilities Either a b will be equal to ed or case two will be a b is less than ed or case three is a b Greater than ed So this one is case two And this one is case three and we'll check for all the three cases if we can establish the ratio Now first case is A So in this case, what happens a b is equal to a b is equal to ed the moment that happens the two triangles become congruent by asa Isn't it? So hence if you see if a b is equal to ed if a b is equal to ed So if you check here, this is e a b ed So a b is equal to ed so then by i by i by by asa I can say triangle a b c is congruent to triangle D e c de f sorry A b c is congruent to de f then hence similarity established They are similar every we know that every congruent figure are similar as well similarity Right, so that's done now. What happens if You know, uh, let's say a b is second case a b is less than de Okay. Now in a b is less than uh De or ed so hence what in this case? What do we do case two? Mark a point on on de so let me mark a point here p P this is p. Okay mark a point p Such that dp is equal to something Dp and let me just finish this. So let's say I'm marking point p and q such that such that Dp is equal to a b and Dq is equal to Ac this I can construct No, don't use trigonometry to prove, you know, uh, see now you are not in a competitive mode guys This is they want you to solve You want they want you to solve the way they have told you Because your answer may be correct. But if it doesn't see please understand the you know, uh, understand from examiner's perspective as well so examiner has been given a Uh marking scheme and he has and to maintain the standards It has been clearly instructed that anything which deviates from marking scheme Will not attract Marks. So if the examiner is very very, you know, under please understand the you know situation he would be Evaluating lots of papers together. So he might not have that much of time to go through each and every new methodology People have done. Yep. Oh, they have adapted then what will happen is they might you know, uh, and even if the A person or the examiner teacher is very, you know, uh, very encouraging in nature. But then Because of the volume he has to handle and to maintain standards He will make sure that the marking scheme is followed. So if you deviate from marking scheme It might not, you know, uh You might not get marked. So hence adhere to cbse rules and regulations To maximize your output. Otherwise, yep, it's not a competitive exam my dear friends where you need to score more from somewhere else someone else Okay, so, um, please understand this. So hence Uh, they are they are going to check whether you know it the way they They have instructed you that's it. There is no other objective. So hence In this case dq is equal to ac and hence join, uh, Pq so I I did this Okay, now if you see this then clearly I would not explain it because you would all know this that triangle abc triangle abc Would be congruent to triangle dpq Triangle dpq triangle dpq correct angle abc is directly, you know Is the is congruent to dpq that means pq is equal to bc Right and why is this this is because of s a s criteria Why because d is equal to a and the other two sides are equal now the moment I say this That means I can write this as so if you see this is let's say x So this is x And if this is y so this is y Right, but then it was already given that e is equal to b. So this was already x here and this was already y here That means Here we get correspondence right By this logic we can establish one fact and that fact is pq is parallel to E f why because if you see angle Dpq happens to be equal to def Okay, so the moment pq becomes parallel to e f my job is done How now you can say that by bpt see dp dp dp by de Is equal to dq by df Right, but dp was equal to ab by construction. So you can say ab by de is equal to ac by Df and that's what you wanted to you know establish. So two sides ratios corresponding side ratios are equal Similarly, so this is one, you know, similarly you have to write there will be two more such ratios You have to write all the other ones like bc by ef Will be equal to ab by de And the last one and hence all the three ratios are also equal Correct when all the three ratios are equal and all the angles were already equal Then hence the two triangles are similar Okay, so this is the proof of Dub triple a that is a a a similarity criterion. So just to repeat you have to draw Dp is equal to ab Then draw dq is equal to ac And then prove that abc is congruent to dpq Dpq like that And the moment you do that you establish that pq is parallel to ef If pq is parallel to ef the ratios are Established right And you write similarly and you repeat it twice and you say that all the three ratios are equal So hence triple a similarity is done Now the best part of triple a similarity is the double a stands. Why because by in this what will you use by angle some property angle some property You know if x plus y plus z is 180 And x dash plus y dash plus z dash is also 180 Let's say x and y and z and x dash y dash z dash are the two three angles of two different triangles and now you know that x is equal to x dash y is equal to y dash Then if you if this if this is true then z has to be equal to z dash Correct So hence if any if all the three angles are equal that means any two will also be equal So hence by angle some property we can establish that You don't need to check the third angle if you check the two angles The third angle By by this property will have to be equal. So hence This is a corollary to the so in a way. This is a corollary To the previous one. Okay. So this is done Now comes triple s. So triple s means if you have all the three sides proportional Then how do you stab this that? The two Triangles are similar So let me again redo it So Actually in the previous case. I didn't take the other case. Okay. You did point it either the other third case was AB is greater than ed here This one Which one Here this one AB is greater than ed but then it was quite similar to the previous case Now instead of drawing pq on def you'll drop pq on abc the only thing changes. You can just write that as well So in the third case you can repeat it and instead of drawing pq on def you drop pq on abc and the process follows same So all the three cases we proved that triple s similarity exists. Okay. Now Coming back to this. So these are the two triangles and it is given that the sides are Proportional that means abc Let me take a minute Yeah, ab this is abc. This is pqr. Oh Yeah, so let's say def def And what is given? It's given that ab upon D e is equal to bc upon E f it is equal to C ac or ca upon fd is it this is given and you have to prove that the two Triangles are Similar, okay, then again, you know the way out again. What do you do put p here? Put q here such that what is this ab? And let me just throw it. Yeah, just a minute. So let me this is yeah So you do this and then what is it? You have ap ab is equal to dp And ac is equal to dq Correct These are the two this is the construction. So and then join pq Okay, now, uh, what do we learn that we know already that ab by de is equal to ab by de is equal to ca by ca by df it is given It is given now ab was equal to dp. So hence you can write dp by de It could be dq by df. You can write this by construction Okay, right then hence what will happen this means this implies by conversor bpt pq is parallel to Here Okay, pq is parallel to here the moment that happens. What does it mean? It means that this angle is x here This angle is x here This angle is x here and this angle is x here I'm sorry not x y it has to be y y why This is why And This is why Okay And uh Now what um, yeah, so it's also given that It's also given that ab by de ab by de Is equal to um ab by de is equal to bc by ef bc by ef right And uh, and so hence what does it what does what do I mean? I mean this let me just erase this Yeah, so let me explain like that. So ab by Yeah, so we established that pq is parallel to ef. We just now need to prove That bc is equal to pq and Then we can say that the this thing has congruent as well Then these two triangles are congruent as well. So how do we prove that? So um So ab by so hence we from here we know that dp by dp by de Is nothing but ab by dp is equal to ab so ab by Uh de correct dp is equal to ab so hence I can write that But ab by de is equal to bc by ef bc by ef right bc by ef Okay, and uh bc by ef okay, that means what? um And just a minute Yeah, how do we prove that uh pq? So pq is equal to dp by just a minute dp by de Is equal to um, what is it dp by de? Yeah, these are parallel these angles are same. So yep dp by de is equal to ab by Be which is equal to bc by bc by ef bc by ef is given okay, and this is also equal to so dp by Be is equal to bc by Oh just a minute it got blank. Hello guys. Are you able to see it? Oh, yeah Yeah, there was some some some glitch. Hello Hello Is the audio clear? Hello. Hello Anyone just oh, yeah, yeah, sorry. So I I got disconnected now. So basically, um, so we established that Uh Wait a minute So pq is parallel to ef and dp Yeah, so hence, uh, where did it go? So what I'm trying to say is when pq is equal to uh, so when x is equal to x and y is y then what do we learn? We learn that triangle dp q is similar to triangle def Why by a a criteria Right both the two angles are same so by a criteria. These are similar that means dp by De is equal to pq by ef Right because they are similar triangles, but we know that dp was equal to ab so ab by de is also equal to bc by ef This is given So this was given and this we proved this we proved so by these two what do we learn? We learn that pq is equal to bc pq is equal to bc right This is what we learn and if pq is equal to bc my friend So all the three corresponding sides become equal That means these two triangles are congruent. Which two triangles are congruent? abc and dcq dpq these two are congruent the moment they are congruent What do we infer that d angle d this angle d is equal to angle a cpct Right now Uh, so when they are congruent. So this is also x And this is y So what do we learn in triangle abc now? So let me write it here in triangle abc And triangle def We have two angles equal Isn't it so x and x is equal to x y is equal to y so by a a criteria a a criteria they are similar okay, so Any doubt in this again? Let me repeat. So in all these similarity proof cases. What do you need to do? You need to establish congruence of you know at least two triangles here Which which triangle so you draw ab is equal to dp ac is equal to dq join pq and try to establish that the two shaded Triangles are congruent. Okay. Now. How did we start we said ab? Uh, we said ab is equal to deep. We said ab by de is equal to ca by df this is given and now by construction ab was equal to dp So we established this and hence we established the parallel line, you know pq is parallel to Ef and hence we could establish the corresponding angles And because these corresponding angles were equal. So we established that these two triangles are equal Similar dpq and def are similar the moment. They are similar Then we established that ab by de Is equal to uh Dp by de but is equal to pq by ef be proved because of similar triangles and by given condition ab by de is equal to bc by Ef this was given so by comparing these two we figured out that pq is equal to bc And that was the missing element the moment we established that bc is equal to pq this one bc is equal to pq what we learn that ab was already equal to Dp and ac was already equal to dq Then the three sides are correspondingly equal that means these two triangles are congruent the moment They are congruent this x comes here. So this becomes x this becomes y And we just needed two angles for proving the similarity. So hence in the smaller abc and the bigger def And the b is equal to e and c equals f and hence by a criteria. We call it similar Okay, now one uh this I'm just skipping, you know, again, this is uh What do you say this is the third one sas again, you have to do the same process Try establishing, you know, uh the congruence And you should be able to do it If not, then let you do just let me know now. Let's you know go towards problem solving as quickly as possible If two triangles are equiangular, then the ratio of the corresponding side is same as the ratio of corresponding medians. Now these are important Theorem again, so they can give you this so quickly. So two The ratio of the correspond two triangles are equiangular angular equiangular means they are Equilateral triangle Right So equiangular means they are equilateral triangle. So hence Two equilateral sorry the diagram doesn't look like an equilateral triangle, but let us assume that abc are equiangular and def Oh, no, but it's one angle of triangle. This is not equi This was oh, where did oh, sorry? I drew I drew it in the wrong card here So two triangles are equiangular then the ratio of the corresponding side is same as the ratio of corresponding medians So hence if you see it's need not to be equiangular actually it is true for any two given triangles. So let me draw the two triangles Like that. So these are the two triangles abc and def And they are let's say a is equal to d b is equal to let's say e And f is equal to f is equal to c Okay, now you draw median d And oh, sorry it is there. So let's say e f g. So a g edge Okay, so they're saying that if they are the angles are corresponding or You know equal then they will the sides will be in the ratio of median as well, which is but True. Well, it is true. Also. Why? Because if you see it's uh, they are equiangular. What does it mean? That means So it need not be equiangular guys. It can be any angle act by the way. So hence let's say ab By de is given to be equal to or since they are, you know all the corresponding Similar right? Yes, if they're similar so all all are equiangular That means corresponding angles are equal that then that means they are similar the moment They are similar then I can write ab by de is equal to bc by e f Excuse me, sir. I'm not here Who's this? What happened Charan tell me So what happens is are you on a wi-fi Charan? Yes, sir. Yes, sir So normally when the you know the the network bandwidth varies this problem arises. So just hang on for a If that persists you can just log out again. No problem Okay, so you will not miss all any ways. Yeah, all this is recorded. So don't worry Yeah, it's fine Yeah, yeah, yeah, so bc by e f is equal to f Sorry, this is ca by fd Okay, this is by similarity now if ag is a median then we know a bc by 2 Is bg so hence This will be equal to bc by 2 divided by e f by 2 as well And bc by 2 is nothing but bg Which is nothing but and b and this thing is e h So hence if you see these two triangles this one And Oh, whatever So this one Oh, never mind. So let's say this one And this one In these two triangles, what happens these two are similar. Why? First of all, the side ratios are same. Here is this side ratios. They are same And there is a included angle also, which is same. So for example here in this case This angle b is angle equal to e so and these two triangles are also similar the shaded ones That means what? a b by a e a b by de sorry So a b by de i can write Will be equal to a g by dh Which are the medians Okay, and similar thing you can do for the corresponding bisector segments as well. So instead of that I uh, so I used to What I used you and in you Never mind. So and the ratio of corresponding sides same as the ratio of the corresponding altitudes all the three can be done very very Similar in similar ways. So hence you have to just prove that The two triangles are similar and you can establish these. Okay, never mind. So let's go to Okay, uh If one angle of a triangle is equal to one angle of another triangle and the bisectors of these equal angles divide the opposite side in same ratio Then the triangles are similar. Now, this is important. This is something which I could be Familiar with okay Yeah, hello, just keep keep keep the mic on mute my friend. Keep the mic on mute. Okay Just yeah, so hence the thing is If one angle of a triangle, so let me just draw This triangle again. Oh, I switched it off just a minute. Oh god 20 switch it on. Yeah, so if one angle for triangle So first of all, let me draw two triangles We use the two triangles and it says a b c This is a and this is d e f. Okay They are saying if one angle of a triangle is equal to one angle of another triangle Let's say a is equal to d And the bisectors of these equal angles divide the opposite side. So let's say this is the bisector This is the bisector and divide the opposite side in the same ratio. What does it mean? So this is if this is x this is x This is why this is why and they are saying let's say this is e f g h they're saying b g the equal And the bisectors of these equal angles divide the opposite side in the same ratio that means it is given that b g upon G c is equal to e h upon h f and you have to say and you have to you know Uh, through that they are similar these two triangles are similar now by angle bisector theorem anyways, we know that if ag is the bisector then Then b g by g c is equal to a b by a c by angle bisector theorem angle bisector And similarly in the second case also E h by h f will be equal to e d by d f Right, but these two have been given to be equal. So hence what will happen? It will become a b by a c is equal to e d by d f okay And uh, what is given is if one angle of a triangle is equal to so hence angle a was anyways equal to a that any way is equal to angle d it was given right so hence by s as similarity criteria we can say that a b c triangle is similar to triangle d e f another point to be highlighted over there is Don't mess with the order of the vertices. So there is correspondence must be there Okay correspondence Must be there right. So hence a must be corresponding to d b must be corresponding to e and c must be corresponding to f you can't say loosely that this is E f d also this will be a wrong statement to make Okay, so do not mess with the order. Okay Now this one is again that was I was trying to highlight is if two sides of a median bisecting one of these sides of a triangle are respectively proportional To the two sides and the corresponding median of another triangle then the triangles are similar very very interesting problem And it might be there as a question maybe so let's say Let's say these are the two triangles one and These two are similar triangles and it is said that this is a b c and d e f Okay, now the thing is if two sides and a median so median, let's say this is the median Let's say this is g And here this is h given is that Uh Proposal to the two sides. Yes. So what what are they saying if two sides and a median? So that means a b by d e is given to be equal to a g by d h Is given to be equal to a c by b f. You have to prove that the two triangles are similar Okay, now this will require some space. So let me do it here Okay, so let me draw these triangles once again. So let's say these are the two similar triangles one So i'm drawing this once again. So the similar triangles are These Oh, sorry, it doesn't look like similar at all. Wait a minute. Oh Yeah, so these are two similar triangles. Let's say a And b and c. This is d. This is e. This is f And it is said that a b And this is sorry not d g and this is h Okay, so what is given given is a b by d e Is equal to a g by d h is equal to a c by d f What do you do to prove to prove? What triangle a b c is similar to triangle d e f Right, this is the given statement now and it's given that a g is median. So it's also given What is what else is given b g is equal to g c and e h is equal to h f These are these are the things which are given now you have to prove that a b c is d f now in this case What you need to do is I will do a construction. Let me just take it away So what construction do you need to do? you have to In such cases, you know, whenever there's a median attached to it What normally you need to do is extend a g Okay to say i and join this Okay, similarly here also extend d h And such that d h i j h j is equal to d h and join f j. So what is the construction? So I'm writing here construction What did we do produce a g? a g is equal to g i Okay, and What next? H d h is equal to h j these two constructions I did and then joined the other two joined Statement again the statement is this Two sides and a median bisecting one of these sides one of these sides. Oh, I'm sorry I have not taken that. Oh, there's a mistake guys. Sorry. So you have to just change this to b instead of a c by d f you This is the difference difference is b c by E f. Yeah, that's the only thing. Yeah, rest everything remains the same. So a b by d e a b by d e is given to be equal to a g by D h is given to be equal to b g by E f. Yeah, thanks for highlighting so This this is yeah Okay, so if two sides and a median bisecting one of these sides of a triangle are respectively proportional to the two sides and the corresponding median of another triangle Then the triangles are similar Okay, so hence What is given that a b by d e again just to reiterate a b by d e is given to be equal To a g by d h. These are the medians and the median is dividing one of the two sides that is b c and e f Is that okay? Hello. Yeah, I hope you understood this now. So hence Uh, the construction remains the same guys. So, uh, what happens is, um Yeah, so hence when I did this if you if you see the triangle a b g and so now consider this triangle and this triangle What happens this side is equal to this side because that those are medians And this is by construction And this is vertically opposite angle. So hence What do we infer we infer that a b g triangle a b g is congruent to triangle I c g Okay by s a s congruency criteria Similarly, if you see in the other set of other other triangle as well triangle d e h is congruent to triangle I f h No, this is j. Sorry j f h j f h by same mechanism That means what a b is equal to c i Right. So this side will be equal to this side and this side will be equal to this side Correct Right. What else so okay, so a b Oh, yeah Again disappears screen gets blank. I think from my side as well. Yeah, never mind. So hence Uh, what do we infer? Um, yeah, we What do we infer we infer from that? Um A b is equal to yeah, so that's that's correct and then Okay, sorry, so, uh So next is how do we go ahead from here? So hence a b g and icj Yeah, indeed it Yeah, so there was some kind of this thing from my side as well. So, you know, uh My hello My uh screen got frozen. Never mind. So let's start. Let's let's go back again. So a b g is icj Uh, sorry, what am I saying? Yeah, so we just established that these two These two are congruent And these two are also congruent Okay, so I hope you can see from here. So hence these two are congruent. So what uh, what does it infer? So a b Uh, a b by D a b by de. Yes. So a b by de. So let me write a b by de Is equal to a c by df a c by df and this is equal to Uh, uh, this was equal to a g by dh. This was given correct Now, uh So I can replace a b by I c right. So if you see this particular thing can be replaced by I c a b can be replaced by ic and de can be replaced by f j Right jf. Let's say so this is ic by jf is equal to Uh, a c. Where is a c? Yeah, a c by df is equal to Yeah, hello Sir shada here Yes shada go ahead Sir, uh in the figure that you've drawn can't you prove that like triangle a b g similar to triangle de H and then say angle b is equal to angle c and then you'll get the similarity, right So how do we prove similarity shada? We don't have many, you know, we don't have any angle or any other, uh, you know, uh So it says a b is the a g like is the median, right? So twice b c like two times b g by two times e f and then two two gets cancelled and then you get Similar to that Wait a minute. You're saying uh, what two times b g is yeah, so which two triangles are you trying to prove as similar? a b g and de h a b g and de uh Okay, yeah, make sense a b g and de h. Okay, they're similar And then angle b equals angle c then you can use s as similarity and prove it Angle b is equal to angle c. Where is it written? Angle b is equal to angle e D is not equal to e. Where is it given that d is equal to e Sir b b b is equal to e. How how how because they are not similar Sir a b g is similar to de Yeah, okay b is equal to prove them to be similar Okay, you have you prove that b okay now And then you take a b a b by de is equal to b c by e f it's given and then angle b equals to angle e s as similarity You get it Okay, so you are saying okay. You are saying triangle a b g Is similar to triangle d h Edge and this you are saying because the three sides are proportional So a b is equal to d h a b a b is Proportional to de sorry a g is proportional to d a sorry. Uh, basically a b by de is equal to a g by d h anyway is given Yeah, so the three by s s s criteria. They are similar. So b is equal to e b is equal to e b is equal to e Then uh in the two this thing s as yeah, looks good. Yeah, good sardar. This can also be done So you don't need to do all this extra stuff So this also can be done But I was why is this crucial or did we uh? Oh my god. Okay. Okay. I understood the problem. Yeah, very good. Yes. So, uh, Okay, now what happens? Let me rephrase this then what if instead of a instead of What uh b g by e f let's say instead of b g by b c by e f instead of b c by e f what happens to the same problem If this is given a b by de is equal to a g by d h is equal to d a c by d f Now the included side is not there You understood. Do you understand the problem? The change problem is instead of these two Proportionality I'm giving you these two now Yeah, so now it now the thing changes Now this is where Probably I was having this thing in mind and you know proving Uh Yeah, so I was I was uh, I didn't change the you know, though I changed in the question But then I was trying to prove when this is given so That becomes the solution. So hence in this case if this is given guys I hope you understood what I'm trying to say It's now given that a b by de Is equal to a g by d h the median ratio And third is a c by d f if this is given If this item is given can you prove that the two triangles are? similar That was the thing so hence in this case it will now become so hence that is how You have to do this construction and then Uh, yeah, this is what I am now going to prove so hence from here I hope that you understood a b by de Is equal to a c by d f This was given a c by d f that's what I was trying to prove see a c by d f and this is equal to a g by a g by d h this was given this was given this was given here So from here, I am replacing a b by ic because I proved them to be congruent. So a b by ic I replaced And d h d uh de I'm replacing by j f here Right. So I got this equal inequality. Uh equality. Sorry. So ic by j f Is equal to a c by d f Uh, I am just multiplying and dividing a g by d h by two. So two a g becomes ai And two d h becomes d j So if you see guys this ratio is equal to this ratio is equal to this ratio Makes you think or you know, you can prove now that triangle a c i Is similar to triangle d f j Okay, so the moment you do this What do we get we get that this angle If this is x then this angle is x Right and If this is why I'm sorry, uh, what is this? It's uh, yeah So this is x And uh, so you get this x is equal to Right. So this these two angles are same Correct. No, similarly you can also prove that this angle if you take the you know, uh, extend the other side You'll also you can prove that this y Is equal to this y Okay, so with the same logic you can prove y is equal to y. So what do we get we get that angle a is equal to angle d Angle a is equal to angle d and that's what we were searching for the moment angle a becomes angle d the other two ratios are same by s a s We can say this also. So there are two problems just to clarify once again one problem said that if a b by d e is equal to a g by b h is equal to b c by E f this was easier to prove which that proved also But the second case was instead of the third thing to be equal We are giving a g by d h is equal to a c by d f Now if that happens then you have to do this construction way Okay, so So let's move ahead because we have to also okay. So time is running out. So let's so this is covered This question number card 22 what you see is what was the next question which we just did Yeah, so let's move ahead now Uh, the ratio of the areas this is very very important and I just told you there will be one marker to marker questions from there So what is this so if you have two triangles again, so two triangles Okay, sorry for these bad drawings. Wait a minute Yeah, so let's say you have two triangles two triangles a b c and d e F if two triangles are there then Then a b by d e d e whole square is equal to b c by e f Whole square is equal to um A c by d f whole square all these squares put together They are all equal to nothing but area of triangle a b c divided by area of triangle d e f Okay, so and it's not equal not only equal to the corresponding sides. They are also equal to Uh corresponding altitude square of the corresponding altitude square of the medians and square of the angle by sectors all are There right, so please remember this So the squares of any two corresponding sides the squares of corresponding altitudes corresponding medians Corresponding by angle by sectors all it will be and one proof. I can just tell you very easily So what is that you drop up a perpendicular from here you drop a perpendicular from here as well Okay, so what happens these two triangles are similar So this angle x is equal to x and this is 90 90. So hence these two also are similar Right when they are similar that means What can you say so area of a b c if now we have to calculate area of a b c is nothing but half into b c into a let's say g and this is h so a g and divided by half into b c Into a h d h Now if you see a g by d h will be equal to If they these two are similar then you can clearly say a g by d h is clearly equal to a c by d f why Because they are similar triangle, you know corresponding sides ratio right now a c by d f already was equal to b c by E f why because the the parent triangles were also similar so hence From here you can write half half gets cancelled So b c upon i'm sorry. This is not b c. This is e f So this is b c by e f Into a g by d h. I can replace a g by d h here with this ratio. So b c by e f again So hence you see b c by e f whole squared Okay, and it will not It will not take much of our pain if you really you know see this is also equal to a g by d h whole square that is the ratio of the altitude And similarly if you're median also you can prove For that as well. So i'm not going into those proves and if at all you are facing any issues you can let me know, okay Okay, uh, okay What is that if the area of two similar triangles are equal then the triangles are congruent Equal and similar triangles congruent understood. This is that If two triangles are similar And their areas are also equal then they are congruent, right? So similarity similarity Plus equal area If these two conditions are met then Congruent triangles Okay, so this is just a statement remember Okay next 25th If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse Then the triangle on both sides of the perpendicular are similar to the whole triangle and also to each other Now this is again a very important theorem and I will explain this Important what does it mean? So if you have a right angle triangle Right angle triangle. So if you have a right angle triangle like that So a b c right and you drop a perpendicular from here like that d So how many similar triangles can you see here? There are Three pairs of similar triangle. So if you see triangle a b c is similar to triangle A d b This is similar. We are similar Okay, this is similar why because a is common and d and b are 90 degree each in both the triangle So by a similarity also Triangle a b c is congruent. Sorry similar to triangle um So c and instead of b you have to write d c d and c d b right Wait a minute. What happened just a minute. So a b c the other one Will be a b c right? We first one is a d b. So the second one is c d b. Okay So c d b means c is common. So and the corresponding to b will be having d and yeah b d c So these this is first one remember This is the second one And the two smaller triangles are also similar to each other and how are they so triangle a b d is similar to triangle Uh triangle b a b d is similar to So d has to be in common b c yeah The big triangle b c d Okay. No, we are not doing proof for proof of Pythagoras theorem I'm just saying there will be there has been question last year on this particular thing that if you drop a perpendicular Okay, we will use this in Pythagoras theorem, but please remember there are three pairs of similar triangles here Okay, so three this particular thing will be used to prove Pythagoras theorem Okay, but then you must know and you know you just do this practice at home as well that how to identify So how do I identify this so I am saying this is one triangle a b d So I have written a you know and then the complete full angle a b a full triangle a b c So if I take these two, so a has to be common in both So hence c a is here a First angle is a then 90 degrees in the first a b c the 90 degree is b so 90 degrees here b And in the second case i'm sorry here in the first case b is here 90 degree and second case 90 degree is d So hence the two is The two angles are equal and now the third one is c and b like that you have to write Okay, so this is how you have to and similarly the other two pair also you can figure it out So in the other two figures also Uh, now in the second case I am taking this this triangle So this triangle and the complete triangle a b c in the c is common So if you see I have first written c At the same position third position and then let's check 90 degree in each case So in the bigger triangle b is 90 in the smaller triangle d is 90 So I wrote b and d like that And then the third the third vertex which is left over I wrote a and b like that Okay, and then same with this also. So there are three pairs of similar triangles And now the Pythagoras theorem so hence I will just uh, you know So hence they will not ask you prove the Pythagoras theorem. They will say This statement so in a right angle triangle the sum of this this this is this they will ask you this To prove and the proof lies in similarity as well. You know already. So let me Use this here itself and prove the theorem as well. So in the first if you take the first one this one This first one first relation For Pythagoras theorem We know that ac square is equal to ab square plus bc square now. I'll tell you how to remember this proof Right, what you can do is You write see, uh, we will go by left right hand side. Okay So A you write you take the first first first similar triangle set of similar triangle abc and adp. So you're writing triangle abc Is similar to triangle adb. So be aware that corresponding angles should be Same Now you write ab here in numerator and then here in denominator then only after cross multiplication It will become ab square Now if you're writing ab from the first one, what will be the corresponding ad so simply write ad And if you're writing ab Here in the second case then here it will be ac. These are the corresponding sides ac And hence if you cross multiply you'll get ab square is equal to ac ad Right Similarly now take the second set of similar triangle. So triangle abc is similar to triangle bdc In this you need to extract bc out. So hence if you're writing bc here Is equal to write bc here also so that after cross multiplication it will become bc square Now if you're writing bc here, so corresponding side will be dc. So write dc here And if this is bc here, that means this one. So the corresponding side will be ac this one So then so hence again, you will get bc square is equal to ac into dc add both these One And two So if you add you'll get ab square Plus bc square is equal to ac is common in both and you'll write right ad plus dc Now ad plus dc if you see is ac So hence it is ac square And it's proved right so pythagoras theorem prove is very important Okay, pythagoras theorem Uh Proof is very important last time in this sample also they have given last few years also they have given this so hence remember the proof step by step Don't mess with the you know the order again. I'm warning you you will be making mistakes here In writing the order. Please be very careful there. Okay And the converse converse is simple What is converse converse is? How to prove the converse that if the Let's say abc So it's given that ac square is equal to ab square plus bc square Okay, prove that Prove that angle b is 90 degree Okay, how to go about it? So you draw another triangle def draw def Such that e is 90 degree you deliberately make it as 90 degree And de ab is equal to de you do And then you do bc equal to ef like that then In the in triangle def what will happen since it is 90 degree so we can write def square is equal to De square plus Ef square by Pythagoras theorem just proved But de was equal to ab so you can replace this by ab square plus bc square ef was bc So but ab square plus bc square is given to be ac square it was given from here That goes on to say that def is equal to ac Right so if def is equal to ac so all the three sides are equal c in these two triangles So that means I can write triangle abc Is congruent to triangle def Hence angle b will be equal to angle e is equal to 90 degrees cpc Okay Yes, so many chats are happening parallely do not engage in chatting and all okay before we end there are few You know more so converts of Pythagoras theorem is done And there are three more slides before we end but before that I want to go to the previous year questions So these are all problems in your ncrt books or you know, so you can always refer to this Uh, this is called apollonius theorem if you if you remember I had You know discuss apollonius theorem in all the classes the sum of the squares of any two sides is equal to So what they are saying is This is a triangle the proof is there in the so please Refer to the proofs as well. So abc the sum of the squares of any two sides So they're saying ab square Plus ac square is equal to twice the square of half of the third side that means And let's say this is d so this is two bd square right Plus ad square This is called apollonius theorem You will not ask by name But they will ask you to Prove this Okay, so by this And similarly the use of apollonius theorem you can prove 29 and 30 slides as well So let me not spend time over there. Those are regular question now Let's straight away come to the previous year cbc question. This was a one marker Okay, so if you see ad is equal to one centimeter given bd is equal to Uh, two centimeter given what is the ratio of areas of triangle right Area of which two triangle ad the top one and the full one And they have they're asking in the opposite ratio the abc you have to write first Okay, so abc triangle abc Triangle abc so hence if you see this is one ad is one and bd is two so hence This is you know, so hence full is three So triangle area of triangle abc divided by area of triangle ade will be nothing but We just learned ab square by ab square by What um ad square Which is nothing but three square by one square Good, so you know this this was one markers for not no big deal. Now. This was the second question. This is previous year paper guys so, uh, a cb is 90 degrees And cd is perpendicular to ab prove that cd square is bd into ad Okay, this is exactly on the thing which we were discussing Uh, just two minutes back. So you have to prove that cd square is equal to bd Into ad Yeah, how to do that. So, you know in this case This is perpendicular. So hence, you know triangle adc is similar to which one adc so you write triangle here and d is 90 so the other triangle c is 90 And a car and c is 90. Okay. So adc So you write bca. Yeah Right, so angle a is equal to angle Sorry angle a must be equal to angle. Wait a minute. Oh, I'm sorry. I'm sorry. Not this Not this you'll have to write Uh, d is common d is 90 degree to both Right. So d is in the middle now a should go to c and c should go to b. Yeah, so this is the thing cdb so adc is No problem. So adc is I will uh, just two minutes guys. I will I just want to show you The model answers as well. Just hold on for two minutes. So if you can adc is similar to cdb. This is what we can achieve. That's what I wanted to Uh, tell you so hence what happens in this case You want to prove cd square so, you know, buy this thing. So you write cd in this case here And here you write cd in the denominator. Why then you will be you'll getting the cd square thing Now if you write cd from here the corresponding side is db. So you have to write db or bd And if you are writing cd here, then here you'll have to write ad so you write ad Correct now cross multiply you'll get cd square Is equal to bd into ad. This is what you have to prove. This was a two marker question last year Okay, so the crucial part was how to identify the two are Similar, so I just used correspondence. So I know d is common in both 90 degrees. So I wrote b simply Then clearly a is not equal to b See the two triangles I'm trying to deal with is adc and the other one is bdc But I just need to establish the correspondence How to establish the correspondence. So d is equal to d anyways now a cannot be so if you're writing a here You can't write b here because a is not equal to b. So hence b comes here automatically And now what is left here c and what is left here? see like that Okay, so hence I established similarity like that So this is how you have to establish. Okay Now this one is very easy one. So I'll just let me yeah, so b and q are the points on the sides C and cb respectively of triangle abc right angle at c prove that so hence the question goes like this So here is the triangle Okay, and let's say this is c is right angle. So a b, okay, and Prove a q p is on ca. So there's a point p and there's a point q Right, and then they're saying Join pb And a q Okay, so you have to prove a q square. So what is a q square? What is a q square a q square is simply a c square direct application of Pythagoras theorem c q square And b p square in the left hand side is nothing but this is three marker question. By the way b p square Is equal to bc square Plus c p square Isn't it now add both of them? You'll get a q square Plus b p square is equal to ac square Plus bc square and I did in brackets plus c q square plus c p square Now if you see c p square plus c q square, I'm sorry See c p square plus c q square is p q square. So hence this becomes ab square and this becomes p q square hence prove easy Right, these are all very very easy problems. Okay see This is Pythagoras theorem four marker last year Pythagoras theorem proof Prove that in the right angle triangle the square of the hypotenuse is equal to the sum of squares of the other two sides We just did it four marker four marks So all the steps will be Yep, and again Application of Pythagoras theorem. This is very easy when you I share the slides you can try at home It was last to last year perpendicular from a on side bc of a triangle abc needs bc at d such that Condition you have to simply use Pythagoras theorem to establish this very very easy Okay, and now so there are a few more questions I wanted to anyway solve but then most of them are application of Pythagoras theorem you can see And easy problems. This is how the model answer looks like See this is the person who has got hundred down hundred in So what are the key points? If you see very neat diagram Needs diagram. So diagram contains a lot of weightage See, you know, whosover it is Given, you know, so they are she's writing Very very, you know What is given? Okay, then all the given is written very neatly Then she has written to prove Then construction the way I explained to you and then followed by proof. So fourth step And then wherever there is a reasoning reasoning has to be mentioned Okay within square brackets, you know, there is no, you know Cut marks or anything just just plain and smooth and wherever she is doing any Let's say if she has to cancel anything one straight cut like that So no doing this. Many people do like this. This is not good presentation becomes very important. See so Numbering of equations see all these are small small steps which Yeah, so very clear and you know Helping the examiner from one point to the other and hence see So use of these signs all those from a and b these are the important things which you can do really You know, so hence neat and so this was another three marker question which she has done. She again given Okay to prove again And she has also mentioned choice one. There are two choices. So she has mentioned choice one also like that You need not mention that but that's fine. See how beautiful and neat diagram it is Yep. See very very, you know, so pencil use of pencil and Yeah So, please, you know, these are these model answers are very important, right by Pythagoras theorem These are, you know, the way they are expressing it in triangle afq angle This you know, all these small small steps are very very vital to score even if you know You need to convert your knowledge into marks Okay, so we know area of triangle is half base into height all expressed so beautifully So wonderful answer presentation very good presentation So they will give you to give them very high on Presentation Okay, and this is one marker. Yeah, so again the area thing which we just solved Okay, so Um Actually, I don't know why they have given them given her mark. This is incorrect So ab by pq. Oh, this was some other question. Sorry my bad. This is some other question some other year So my bad. So but then similar type of questions are being asked every year Okay So this is how one marker here if you see they have not mentioned any reason so reason is not many here No reason has been mentioned that square of Sides ratio sides is equal to square of the area Okay, so hence you can you know in one marker you can afford not to write But for all other things you have to be very very You know specific thorough neat and Methodical so if you skip steps, probably they will You know cut marks Yeah, no, it's not scam. It's not scam. It's some other question some other question. So my bad Anyways, so I hope the session was useful to you now this session the full Okay, this thing will be record, you know, it's already been recorded. So it will be uploaded on youtube and you can The next session is tomorrow. So be there For I believe there is chemistry class if I'm not wrong. So I hope You know, so after this will be also posting the previous year question word worksheet So I would urge you that now that the you know, you are free from any other school work or exam Please solve them one by one and complete your preparation. So right all the best guys. Thanks a lot for attending Bye-bye. Good night