 Triangle chapters if you see is categorized or you can actually see there are three broad category of topics So the first part we dealt with something called basic proportionality theorem Basic proportionality proportionality theorem then we have This is also called Thales theorem. So if you if you know we have discussed this in the class. So Thales theorem so this is the second part of it and Then we had similarity criteria. So if you see we had similarity of triangles similarity of triangles and the criteria related to that and then third part is Pythagoras theorem and its application Pythagoras theorem the age-old Pythagoras theorem and And the problems related to that. So this is in a nutshell problems related to these Concepts, okay, so in your board paper whenever you get to see Any triangle or you know geometry related question. So if you know that in your boards board syllabus geometry Has two subparts primarily where one is questions related to triangles would be there and second one is circles and There will be questions there could be questions which will be requiring you to know both the concepts and The same question could you could have applications of both the Both the subtopic soon. So today we are going to discuss What all did we study in triangles and how to approach problems whenever you're going to face these and let's say in your board paper or mock papers In the days to come. So whenever you you know see a geometry problem Then you know that there these are the concepts which are going to be Implemented there now questions related to ratio. So hence these are the things. These are the catch words You have to you have to be sure of so ratios then there is you know, if you see parallel lines Parallel lines in in problems. Let's say in a given figure. There is you know, there's a figure and there is a triangle There are two triangles like that and then one parallel line is this another parallel line is this like that if you see such kind of You know figures. Let's say this is a b c and d And e f g let's say this is a figure given and something related to this figure There is a question which involves ratios parallel lines or maybe let's say squared square terms Square terms, this is how you should be knowing so square terms. Let's say they'll ask you Let's say a c square is equal to b c into cd and things like that Whatever so whenever you see such kind of relations to be proved You know that this has got to do something with Triangles and either of these three either of these three concepts either Thales theorem similarity of triangles or Pythagoras theorem would be Applicable so keep that in your mind now. Let's go and revise all the basic theorems and results which you Must be having in your mind before attempting all these questions related to triangles What we have done is we have compiled all the major theorems which are there and the results which are there and which are useful for triangles chapter and Later on you can you know get these slides for your own you know reference and we have also planned separate sessions on problem solving which will be Which are you know which are being recorded right now and they will be uploaded on your on our YouTube channel So you can always reference the problems Offline so let's begin the summary. So this is the first slide. So basically there are 30 slides We'll try and you know Complete as many slides as possible and if possible we are going to complete the entire derivation today itself So first slide is two figures having the same shape but not necessarily the same size are called similar figures We'll run through these concepts. So two circles are always similar. Why because if we have two circles we have You know, they are similar. We know that two circles are always similar so All congruent all congruent figures are similar, but the converse is not true. That means if there are two congruent Congruent triangles or any two congruent geometric figure, they will be similar But the vice versa is not true. Why what is congruence congruence means if all the angles and all the sides corresponding angles and corresponding sides must be equal Must be equal. This is My congruence So this side is equal to this side. This side is equal to this side and this side is equal to this side And all the corresponding angles are equal then we say that The two figures are Two figures are congruent, but For similarity, you just need Scaled one one geometrical figure scale down or scaled up vis-a-vis or with respect to the other one, right? So that's that's similarity. So hence all congruent figures are similar, but the converse is not true So these two triangles might be similar, but they are not congruent. Why because the corresponding sides Corresponding angles are are same are same but corresponding sides are not equal corresponding sides are not equal So hence if you see this is a b c a b c and this is d e f This is d e f then a b is not equal to Not equal to d e Okay, so though though they are similar, but they are not Converse so this is point number one next is Yeah, two polygons have the same number of sides are similar if so this is the criteria for similarity if you see So this is a this is the criteria for similarity to polygons having the same number of sides are similar If they are corresponding angles are equal and their corresponding sides are proportional in the city that is in the same ratio So we'll start with our triangle our triangles only so let's say this is a BC again same thing a b c and d e f Then these two triangles are similar These two triangles are similar only one condition is first of all all the corresponding angle should be equal So angle a must be equal to angle d Angle b must be equal to angle e and angle c must be equal to angle f This is criteria number one and criteria number two is a b upon d e is equal to b c upon e f is Equal to c a upon f d these is criteria number one criteria number two so both the criteria are met then to given this is I have shown shown it for triangles, but it is You know same it holds similarly for any other polygon as well. So corresponding angles must be same and corresponding Corresponding angles must sides must be proportional. This is a criteria for similarity next So let's go to the next slide. It says 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 this is nothing but Heli's theorem if you if you remember this is nothing, but if a line is drawn parallel to a Triangle, so let's have a triangle So let's say I have a triangle here triangle and This is ABC this is Thaley's theorem. So or basic proportionality cell theorem. So this can also be denounced. This is my basic proportionality proportionality Basic proportionality theorem right or it's also called Heli's theorem whenever while solving the problem You know, I have seen that many of you write just BPT. That's okay But then You whenever you are using the theorem for the first time you have you should it's a good practice to always mention the acronym Which you are going to use later on. So, you know, this is a common practice Now what does it say if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points That means let's say if I draw a line Like that then the other two sides divided the same ratio. Let's say this is D and E So we know by BPT. What does it what does it say AD upon DB is Equal to AE upon EC AE upon EC and I'm not going to the proofs of it, but I'll just tell you the you know The basic way of proving it. So basically have to join DE and drop perpendicular from E to let's say This is perpendicular drop from point E on side AB, which is let's say D and then if you remember we used the concept of Try area of triangles to you know, you have to you have to join this as well. So And then you have to drop up a particular like that So if you if you see what was the you know, I'll just give a brief Hint of what exactly the proof was about. So basically we found out that we took first of all triangle ratio of triangle area of triangle a ED over triangle be DE now these two triangles this is area of we are trying to find out the ratio of the area now in these two triangles base is You know the height is same ED is same and hence this was equal to nothing, but AD upon AD upon DB why because this will be similar The triangle area was nothing but half into base into height height is same. So it is reduced to ratio of the Ratio of only the base is similarly. You have to take the another other triangle. So that is area of triangle ADE area of triangle ADE divided by area of triangle DCE DCE here also if you see DCE here also the ratio will be nothing but a E upon a E upon EC why Because the heights are same So it will be in the ratio of their bases. Now if you see triangle ADE a Or sorry triangle DCE DCE DCE is this triangle this area this area is This area is equal to the area of this area Why because the base is same D is the base for same and they are between the same parallel line So hence from that logic area of triangle BDE is equal to area of triangle DCE Hence these two ratio become become equal So hence that was that is how it was it was Proof 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. So again if you see This is our triangle, let's say we draw a triangle. So if we have a triangle like that and We are saying that there's a if if if there's a line Which divides this triangle ABC the sides of this triangle in such a way that that AD upon DB is equal to a E upon EC EC then we say that we declare that DE is parallel to BC DE is parallel to BC. This is converse of Converse Converse of BPD. Okay, so keep that in mind now How did we prove that we used something called contradiction? I'm not going again into deeper details Contradiction will be used contradiction means let us say that Given this ratio is there. Let us say that D is not parallel to BC and we say that let another point D e dash be there on AC such that D e dash is Parallel to BC then by can buy our BPD we can say that AD upon DB is equal to a e dash upon E dash C why because this is just what we proved that if D dash is parallel to BC Then AD upon DB will be equal to a e dash is upon e dash C but if you see the given is this this was given so that means from this relation and this Relation we can say that a e dash by e dash C is equal to a e upon e C Isn't it now? That's the if that's the case that means there is to there are two points on AC which divide AC in the same ratio. That's not possible and Hence we say that e dash must be Equal to e that means e dash must coincide with e and hence and hence we say that D e will be parallel to BC. That's how we proved converse of BPD. Okay enough now So let's go to the second next one the next Slide says the internal bisector. This is called internal bisector theorem So these are the you know basic theorems. We must remember the internal bisector of an angle of a triangle So let's say if there is a triangle like that ABC is this ABC is a triangle and there's internal bisector of an angle. Let's say angle a is being bisected Such that this is D. So a angle. This is if this is angle theta This is uncle theta. Let's say the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of contain of in the ratio of the sides containing the angle that means if ad is Ad is internal bisector internal bisector bisector of angle BAC If that is true, then we know it is BD upon DC is equal to AB upon AC So this is these are the sides containing the angle AB and A so angle A is being bisected So hence AB upon AC is equal to BD upon DC So hence you remember like that a B upon AC is equal DC. Yeah, this is angle bisector theorem Again, so how do you prove it? So if you remember to prove this you have to extend AB You know in this direction such that AC is equal to let's say this point is E So basically we have to prove that a AB by AE is equal to BD by DC Then our job is done. That means somehow you have to prove that ad is parallel to CE if this angle is equal or These these two sides are these two sides are equal or alternatively what you can do is you can reuse CE in you can make CE parallel to ad and let let BA extended Join CE at this point Okay, so using these so this is how You and then you use an DPT you can prove that AB is equal to or AB by AC is equal to BD by DC theorem So next term is if a line through one vertex of a triangle divides the opposite side in the ratio of the two sides Then the line bisects the angle at the vertex. So basically it's converse of the previous theorem So if you see this is nothing but converse Converse of the previous theorem converse of Previous theorem. What was that and what does it say? So basically if again you have a line joining let's say a a b c and then and then You have let's say this is BD Okay, so if a B is the point on the C such that it's given that BD by BD by DC is equal to a B by AC then Then angle BAT will be equal to angle D a C This is what the theorem suggests So next is the line drawn from the midpoint of one side of a triangle is parallel to one of the side And it by six the third side it is a direct fallout of Thales theorem. So let's let's again understand what exactly is this So there is a triangle There's a triangle and this is a BC and We say that line drawn parallel from the midpoint of one side. So let's say this is E is the midpoint E is the midpoint of one side of a triangle and It's parallel to another side, let's say we draw a line EF or let it be D instead of this being E let it be D Okay, so let us let us say this is point D and it's given that D is midpoint D is midpoint of AB Okay, D is midpoint of AB such so what does it mean AD is equal to DB? And then we are drawing a line parallel to line parallel to BC So clearly it did, you know, you can understand from this direct result of Thales theorem If DE is parallel to BC, then we know that AD by DB will be equal to AE by EC right by Thales theorem and hence the first part of this thing is proved So hence AD by DB anyways is one. So hence AE by EC is equal to one and Hence E is the midpoint as well E is the midpoint Okay, the next part of this is is that DE is Now once E is midpoint so you say the other part is DE is half of BC Right, and if you remember, I'll just again, I'll give you the hint why how it was done So you'd have expended DE to EF DE to EF and such that DE is equal to EF then what will happen? You will say that If DE is equal to EF, AE is equal to EC anyway, then this angle is equal to this angle. So triangle AD E is congruent to triangle Triangle CFE Is it it? Once that is done, that means this angle is equal to this angle And hence this angle is equal to this angle So basically by this you will so basically DVC is equal to DF DFC these two angles are same and what else if you see So basically DF CV can be proven to be a parallelogram. So this is a parallelogram how? because This angle is is equal to this angle that means CF is CF is parallel to AD which is parallel to AB so CF is parallel to AB and AD is equal to DB is equal to CF why because by congruence we proved AD is equal to CF or FC and AD was anyway is equal to DB that means DB is equal to CF and we just proved that AB is parallel to CF that means DFCB is a parallelogram that means DF is equal to BC and DF was nothing but 2DE Is equal to BC so DE DE will be equal to half the external bisector of an angle of a triangle divides the opposite sides externally in the ratio of the sides containing the Angle so what does it mean? This means simply this similar to internal bisector theorem We have external bisector theorem. So let's say ABC is a triangle and B a is extended. Okay, and let's say this is my So don't go on the you know scale. It's not to scale actually Yeah, no worries. So let's say this theta is equal to this theta and let's say this is a this is D Okay, so D is the AD AD is external is external bisector External bisector of angle C A and let's say this point is E C A E Okay, so if that is so then we know that AB upon AC AB upon AC is equal to is equal to BD upon DC Okay, that's what It will be right. So how to remember this so So you start with AB and AC and hence you also start with from this point B So let's say AB is the first side AB and then B is the B is the Starting point of the side on which the point B lies Then you first stretch out from B to D and from D to C like that, right? So hence AB upon AC will be equal to BD upon DC like that Okay, so this is external and external bisector theorem Okay, let's move on. So this is done cards a 8-7 is done. Yeah, so internal angle bisector theorem external bisector theorem midpoint theorem now The line joining the two midpoints of the sides of a triangle is parallel to the third side again you know result of Converse of Converse of Thales theorem so you can prove this by Converse of theories from ABC is a side and D is the midpoint of D is the midpoint of D is the midpoint of AB and E is the midpoint of midpoint of AC then D is parallel to BC D is parallel to BC converse of this can be proved by converse of Thales theorem how so ad by db so clearly ad by db is equal to a e by e c Is equal to 1 by 1 so hence Since the ratios are same by converse of Thales theorem D will be parallel to BC, okay next so let's go on to Yeah, diagonals of a trapezium divide each other proportionally Okay, so let us understand this. What is this Diagonals of a trapezium. So what is a trapezium trapezium is a quadrilateral having two sides Parallel and the other two sides need not be equal. Okay, they're equal. We call them isosceles trapezium abcd is a trapezium abcd is a trapezium and the diagonals of a trapezium. So let's say diagonal ab and diagonal ac Okay, so this let's say this point is o. So what Does this theorem say the diagonals of a trapezium that means od upon ob Is equal to oc upon uh oa Okay, so they are dividing each other proportionally And again very easy to remember the proof what you do is you just draw a line parallel from o onto And join it with ad and let's this point be let this point be e Okay, so uh, and then you can also Yeah, so from here you will get From here, what will you get you will get? uh in triangle in triangle d ab In triangle d ab what will happen by theories again by bpt By bpt Since eo is parallel to ab then de upon ea Is equal to do upon ob Mind you guys. I'm not writing all the statements because again This is a division session. So we'll try to give you as many hints and Um, you know, you have already gone through it. I'm just trying to recap everything. So hence You should not be writing like these When there is a question on proof you should not adopt this methodology for all though Questions previous yes questions. We have we are doing a detailed model answers Writing techniques. So hence you can always visit those videos to see how the answer should be written. This is just to recap Whatever we have learned so that you can retain these during your exams and you can apply this So d upon ea is equal to do upon ob by therese serum and by therese serum again You in in triangle this was in triangle d ab Similarly, you can also take triangle Yeah And then now you can try you can take a triangle adc adc in this also again a e upon ed is equal to a o upon oc a o upon oc so if you see These two using these two Using these two you will see and you have to see for example here a e upon ed is there here it is The reciprocal of that ratio and hence If your reciprocated it will become ob by do and if you equate them You'll get a o this this particular relationship Yeah, so it falls out directly from here, right? So that's how you can prove this theorem Next slide says if the diagonals of a quadrilateral divide each other proportionally then it is a trapezium again converse So, let me yeah, so this is the converse of the previous theorem itself So converse again. This is the converse of The previous slide. So what does it say it says that let's say there is a trapezium Okay, let's say there is a trapezium a b c d And if they are let's say there's a diagonal There is a diagonal which is like that and Yeah, and this is o and is given that a o by oc Is equal to let's say bo by od then Then oh, sorry, this is a quadrilateral. It's not a trapezium. I'm sorry So if there's a there is a quadrilateral and the diagonals are bisectors not bisecting Dividing each other proportionally. So this is how they will divide Proportionally then then we have to prove that Then we have to prove that a b c d z trapezium a b c d is a trapezium So how to do that? Okay, so let us say that we are we we draw a line parallel to So again, you can take this as you Yeah, you take you find a point e such that If you find a point e on ad you can always find a point e on ad such that A e by ed is equal to a o by oc Okay, there's no problem So you can find a point e on ad such that a e upon ed is equal to a o upon oc And hence which is equal to bo upon od because this was given anyways. Yeah, so that means You have to take first this combination which combination. So these two You have to take these two and then if you can if you take these two ratio then by converse of By converse of dp t you can say in triangle adc In triangle so from this what follows in triangle adc In triangle adc what happens? e o becomes parallel to dc Isn't it by converse of Thales theorem and then if you take these two together that is these two together not not these uh, basically a e by Ed is equal to bo upon od if you take these two ratios together then what will happen You will see that oe is parallel to ab oe is uh, yeah oe is parallel to ab and then you can combine these two to get what You can get ab parallel to dc that's what if you prove this that means that means abcd the quadrilateral with two parallel opposite parallel sides that means it is a trapezium okay Next yeah any line parallel to the parallel sides of a trapezium divides the non-parallel sides proportional now instead of diagonals Now they're talking about let's say if there is a trapezium And any line parallel to the parallel side So let's say if I draw a parallel line like that then they're saying if this is a This is b. This is c and d then they're saying let's say this is d and sorry. It's not d. This is Wait a minute. Yeah This is e and this is f and it's given that E f is parallel to ad is parallel to bc E f is parallel to ad one side and it's parallel to bc then they're saying that a e d by e b Is equal to b f by fc Now how to do that so you can do it multiple ways and you know one is you produce the Two sides of two non-parallel sides to meet at let's say oh Okay from here from here if you see you you get what? Again, you know I said that I will not be covering proof but then quickly if it is possible Let me just try and cover as many as possible. So oh a So in triangle, uh o e f again o e f. What can I write by bpt? In this triangle o e f ad is a parallel line So I can say o a by e a is equal to o d by d f Isn't it similarly in triangle o bc we can say o a by o b is equal to Or rather o a by o a by a b just a minute. Um Just a minute. I'll delete Yeah o a by a b o a by a b is equal to o d by dc Okay, now what you do is you divide one by two One by two if you divide one by two you will get a b by e a a b by a b by e a Is equal to cd or dc whichever way dc by df dc by df Isn't it now I am writing in this side and then subtract one from both sides you will get a b by e a minus one is equal to cd by df minus one And if you if you solve it you'll get a b minus ea on a b minus ea on the left hand side, which is nothing but so I'm just writing it here. So it's nothing but a b minus ea So if you take the lcm and all you'll solve it you'll get a b minus ea upon ea Now a b minus ea is b e so you'll get on the left hand side b e upon ea Be upon ea and on the right side cd minus df is cf so you'll get on the right side cf upon cf on df so if you see Friends, this is nothing but this Or the reciprocal of that right so hence you proved this as well Again if three or more parallel lines are intersected by two transversals That means if let's say there are two or more parallel lines. So what does it mean? So let's say this is a b cd ef Okay Okay, a b cd ef then they are saying a c upon c e is equal to bd upon df Yeah, so this side upon this side or intercept is equal to this side on this side, which is again Very easy to prove how you produce this You produce this and then again take triangle a cd first and then take triangle a e f with a b as the parallel line a b parallel to cd So you first take this triangle which one a cd From that you will get let's say this point is o so from that you'll get oa upon ac is equal to ob upon bd and then take triangle o c or o a e Sorry o e f in that you'll get oa upon a e is equal to ob upon bf and from whatever we just Uh Whatever we just proved In the previous card you can use the same to prove this. I hope this should not be a big challenge for you guys Okay, same again so far. We have been doing applications of bpt Okay, and just to add add here. I will also tell you that in bpt There are multiple corollaries. So hence you must be knowing and you must you can always quote this in your exams By proving sums. So let's say abc is the triangle and this is de is parallel to So de is parallel to BC then all are true. What all ad upon db is equal to a e upon ec Also ad upon ab Is equal to a e upon ac Also db upon ab Is equal to ec upon ac all these three are True you can do it by adding plus one minus one and things like that So there's a question that in card 13. Okay, let me go to card 13. So card 13 says wait a minute So in card 13 Yeah, so in card 13, can we just can we just construction and diagonal and prove it? Yes, you can do that as well. There is no problem. So you can do that as well Whichever way you want to do this that both are there would be multiple ways of proving the same question But question critica is asking is this let me share it with all other people So there are three. Let me just now delete this and in this card. She's saying if there are three lines This is not parallel obviously it doesn't look like so this is let's say there are three lines parallel And there are two transversals like that So what she's saying is can I join these two? and prove You can definitely prove how what is the proof? So let's say all of you would be knowing it a CD and e f So again by application of vpt. Let's say this point is oh so a c upon c e is equal to b o upon o e And in again integral b e f it will be b o upon o e is equal to b d upon t f Both is this is also valid proof. I hope I clear this Next let's go on to our slides back So so you could also do if you represent using the time. Yes. Yes param correct. You're correct. So you can use You can use thanks for adding to that. Okay. So card 13 is done card 14 is done Oh Oh, sorry Okay, okay. Okay. Yes. I was answering params question. So critica you you're this thing also be done in the same way Which I just proved for card 14. So the drawing which I did here was for part 14. So that's a correction Next is okay. Now comes the second part. What did we discuss? We discussed these three parts, isn't it? So triangles basic proportionality theorem so far we learned around 14 14 theorems or let's say you know 14 piece of information around basic proportionality theorem it included midpoint theorem it included internal and external bisectors theorem So all should be there on your fingertips. Now we are entering the second realm and that's similarity of triangles. Let's see how So first is the criteria for similarity. So we learned criteria for similarity in some slide back and we said that corresponding angles must be equal and corresponding sites must be proportional, isn't it? So Yeah, so we for two criteria for similarity Which has to be true. It's another thing that we can we need not check everything for its validity, but here here So we have some ways to reduce our Checking criteria, but then these will definitely be true always what all Corresponding angles corresponding Angles Must be equal. This is point number one And two is corresponding Sides must be proportional sides must be Must be proportional, isn't it? proportional So these are the two criteria now you don't need to check each and every criteria for uh, you know checking the valid all It's a similarity of two triangles. So hence we have some, you know, we you can eliminate Your requirements some so not all the rest there. So hence there is six criteria, right? There are three angles and three Proportional sides. Okay, so let me say So actually if you see you know that even if Three all the three angles are same. Let's say this angle a is equal to t And b is b is equal to e And c is equal to f Then also the two two Triangles are similar and I will just try and do one proof for you arrest. You can always You can use Uh, a similar method to prove all these criteria though the criteria Usually they do not ask proving problems or criteria proving problems In uh in boards, but you must know Okay, so hence if you see what do I do I will Select a point e dash on a b such that a e dash is equal to A e dash is equal to d e. So I select a point e dash such that A e dash is equal to d e and then I select another point f dash on On this thing A c such that a f dash is equal to D f and then what do I do I join these? So when I when you join this what will happen if you see triangle clearly triangle a e dash f dash is congruent to triangle d e f by s a s congruency criteria, isn't it because angle a e dash this side is equal to this side And let me use another color. So that becomes easier for you to understand. Okay Okay, and this side is equal to this side And boss this angle is equal to this angle. So hence s a s criteria. These two triangles are Congruent so these two triangles are congruent that means corresponding Angles must be same that mean easy angle is equal to angle e dash and angle f dash is equal to angle f Is it but my friend b is equal to e and e is equal to e dash that means b is equal to e So if b is equal to e dash that means line e dash f dash is parallel to bc Right and if e dash f dash is parallel to bc then by using bpt that is basic proportionality theorem You can always say a e dash. So hence the conclusion would be a e dash by e dash b Is equal to a f dash by f dash c and hence you can and a e dash is nothing but d e so hence And and and the corollary to it will be nothing but a e dash by a b is equal to a f dash by Ac and now you can replace a e dash by what d e and a f dash by d f. So hence So, uh, you know, uh, we we now say that even the two sides are Two sides are proportional Two sides are proportional now, you can repeat the same process for angle b and then angle c and you will see that All such sides are All the corresponding sides are also proportional. This is just one Case there will be three such cases in all the three cases. You can prove that the corresponding sides are Proportional so this is the proof for let's say triple a similarity criteria. So that means if in two triangles Corresponding angles are equal then the triangles are similar Fair enough. So let's move on to our next slide Okay, so the next thing is a similarity criterion. I don't need to go much into deeper details. Why because are Because of and you know angles some property of a triangle they say that you don't actually need All the three angles to be checked if two angles are checked to be equal corresponding angles to be equal Then you're done Then the two triangles are similar a b c and this is d d f. Let's say d d f and Angle a is equal to angle d and angle b is equal to angle e then automatically by angle some property what by angle Some property which says that some of the three angles of a triangle is 180 degree So by angle some property angle c is equal to angle f and then by the previous slide now We have a a a similarity and hence the two triangles would be similar again So hence no worries. This is clear to everyone. I believe let's move on to the next slide Okay, triple a similarity criterion now they say that if If again, so let me use this triangle only. Okay, so let's say It says that triple a triple s is what if the three sides are Sides are proportional that means a v a by d e is equal to bc by e f is equal to C a upon f d Right then Then then triangle a b c Is similar to triangle d e f Mind you guys while writing similarity, please be very careful in writing the corresponding vertices against each other So triangle a b c need not be similar to triangle e d sorry F e d or just let me write it properly. So need not be similar to F e d so because Corresponding angle a and f are not same a and d were same friends a and d must be written in the same Same order. Okay. So all all the vertices must be in the same order of Equality that has to be taken care of Okay, so let's move on to Next slide slide Yeah Now if one angle of a triangle is equal to one angle of another triangle and the sides including these sides are in the same ratio Then the triangles are similar. This is nothing but s a s. This is nothing but wait This is nothing but S a s So s a s s s right so now Wait, yeah if One angle of a triangle is this should be a triangle. Sorry. So if one angle if one angle If one angle of a triangle is equal to one angle of another triangle And this and the side it's given that Upon d e is equal to a c upon d f Right if this is happening if this happens if this happens Then we say that triangle a b c is similar to triangle d Tangle d b f Right, this is nothing but s a s similarity Now so just to summarize before we proceed further. So how many similarity criteria we learned one is One is a a a This is once then second is a a so these are these are mostly same You know third is s s s And fourth is s a s. These are the four similarity criteria. All right, so please be Mindful that one of whenever you are proving similar triangles one of these criteria has to be Used okay fair enough. Let's move ahead and now see applications of similar triangles If you see you know application of similarity application of similar triangles and we'll see how This these applications turn out to or let's say Produce these results. So what is that first of all T? I'll try and prove one and then rest you know Doesn't take much of a effort to prove the next ones. So if two triangles are equally angular that means They have Same angles right if two triangles are equally angular Then the ratio of the corresponding side is same as the ratio of corresponding mediums We are on card number 19 and then it says if two triangles are equi angular Then the ratio of this corresponding sides is same As the ratio of corresponding mediums. Okay equi angular that means we have to take equi angular is nothing but Equilateral triangle. So let's say Okay, so hence let us take two Equilateral triangle Now the question is let's say if you have this as a DNC and this is DEF Then what what are they saying if two triangles are equi angular then the ratio of the corresponding side Is the same as the ratio of corresponding medium. So very okay Okay, so the ratio of the corresponding sides is same as the ratio of corresponding mediums So let me just draw a proper median here So this is the median So median what is the median median divides the opposite side You two equal parts. Okay, so let's say this is G and h okay Now it says that if Two so hence a b It is given that a b by de Uh, obviously if they're if The two Triangles are similar or equilateral or equi angular then a b by de is equal to bc by EF is equal to Fb by ac Isn't it now so Just Yeah, so hence now if this is so then This is equal to a g by D h sorry. Yeah, D. This is g h okay Okay, and hence the other medians as well. So let's say if this is I g h i and let's say this is j then Also equal to me i by e j Okay, ac by f d. Okay, so did I oh sorry I I wrote it the other way around Thanks prajna for connect correcting. So it is a b by de a b by de is equal to bc by e f is equal to ca I'm sorry. I'm sorry. This is wrong. This is wrong and this has to be Corrected thanks for correcting and this is ca upon f d. Okay now again How do you prove? So if these are equi angular then I don't you know, uh this to This two angle is same This two angle is same as 60 degrees And bg is bg is nothing but half of bc and e h is equal to half of hf because ag and D h are median. So hence if you see by angle side angle side again abg Triangle abg triangle abg. Let me write it down triangle abg is similar to triangle de edge Why because ab by Ed Is equal to half of bc by half of E f that is nothing but bg and bg upon e h and then angle abg is equal to angle de h Isn't it so hence by what similarity criteria by s a s These two triangles are similar hence the corresponding sides will be equal to ag upon de h Okay, similarly corresponding sides is same as the ratio corresponding angle bisector segments What is angles bisector? So in case of equal angle inqui angular Triangles angle bisectors altitudes also center in center or all sorry not centers Altitudes medians Um your internal angle bisector and perpendicular bisectors all are same Okay, so hence this will be true for all of them This will be true for all of them. You can try so what did we learn one is median? What is median for your this thing median is nothing but line segment joining vertex to vertex to opposite side Opposite side of the triangle. This is Opposite side midpoint opposite midpoint of already right midpoint Midpoint of opposite side of the triangle. This is median then Altitude what is altitude? Altitude altitude altitude is nothing but perpendicular perpendicular from One vertex One vertex to the other To the other side opposite side this is my Altitude then there is something called angle bisectors segment angle bisectors. It's very clear So angle bisectors are nothing but line segment bisecting the angles of The triangle and then there is perpendicular bisectors perpendicular bisectors. These are four The optical you know elements which we study so median altitude angle bisectors and perpendicular bisectors so angle bisector is nothing but this is angle bisector Let's say this is the triangle. So this will be my angle bisector. So this angle is this angle And this will be my perpendicular bisector, right? So this side is equal to this side Okay, so you know this Let's move on next if one angle of a triangle is equal to one angle of another triangle and the bisector of these angles divide the opposite sides in the same ratio So then the triangles are similar. What does it mean? Let's see So if one angle for triangle, so let's say two angles are there two triangles are there one Right a b c it says what a b c d e f two triangles are there. What is given? It's given that angle a is equal to angle d Okay, one angle of a triangle is equal to one angle of another triangle and the bisector of these equal angles So let's say so you have these bisectors of these angles. So what what does it mean? This angle is equal to this angle This angle is equal to the same bisectors of these equal angles divide the opposite side In the same ratio. So it's nothing but the converse of Converse of our internal bisector theorem So if one angle of a triangle is equal to one angle of another triangle Okay a is equal to d and the bisectors of these equal angles that means this You know a b c And then this is a d e f g. Let's say and this is h Okay, so it is saying that a a g is bisecting So a g is the angle bisector and it is dividing the opposite side in the same ratio. That means it's given that b g by g c is equal to e h by h f Then the two triangles are similar So if i'm not wrong, it may not always work for perpendicular bisectors as they don't pass through the opposite person Oh in card number 19 you're talking about just let me see. What was that? Uh, okay, there's a doubt So it says if you try to say equal angle, then the ratio of the corresponding side is same as the ratio of the corresponding altitudes Yeah, yeah, that's true because they are yeah, so in the case of equal equi angle time. I was just Mentioning about all the four elements because those four elements are usually so I was just trying to recap The four elements of the triangle which are usually, you know utilized in problem solving. So yes, it will not hold true for The perpendicular bisectors as it is rightly pointed out that perpendicular bisectors will not Pass through the vertices very true correct. Thanks for correcting So card 22 is this and we were explaining this to you Yeah, so hence it's given that angle a is equal to angle d and bg By gc is equal to e h by h f. It is given then you have to prove that a or basically Triangle a b c so hence if that is true these two criteria are fulfilled then triangle triangle a b c a b c is similar to triangle d e f Right now So hence what did we learn from internal bisector theorem? If you remember bg by gc is equal to a b by ac Internal bisector theorem says that in a triangle if there's an internal bisector of an angle Let's say ag is the internal bisector then a b by ac will be equal to bg by gc That's what we learned in that internal bisector theorem. So hence a b by ac So here a b by ac is equal to bg by gc Will be equal to similarly here will be equal to d e by d f for d f So this is d e by d f by d f Now b a b by a b by ac is equal to d e by d f and angle a is equal to angle d Right already given one angle is given from here angle a is equal to angle d. So by s as criteria side angle side criteria those two triangles are similar Okay, so this is Also done So if two sides and a medium bisecting one of these sides of triangle are respectively proportional to the two sides And the corresponding medium of another triangle then the triangles are similar. Let me explain looks little complex while reading it So what does it say? So hence you should do like this two sides, okay, so two sides of and a median Bisecting one of these sides of a triangle. So let's first make one triangle one triangle is like that So let's say a b and c. Let's say Two sides. So let's say let's pick up a b and ac these two sides and one of these and A median bisecting one of these sides. So let me pick up this ac And let's say this point is D okay, so Triangle are respectively Proportional to the two sides. So we'll have to have another triangle. Let's say another triangle like that It's proportional to let's say p q r and s Right. So it is saying a b by a b by ac Is equal to If two sides And a median i'm not sure i'm i'm sorry. This is not a b by ac. It's basically a b by a b by p q a b by p q is equal to a c by p r And is equal to b d by q s If this is so then Triangle a b c is similar to triangle p q r This is my problem. So hence again Again, how to read two sides and a median bisecting one of these sides. So I pick two sides a b and b c And median bisecting one of these two sides one of these two sides is Let's say ac Pick d as the midpoint. So b d becomes the median. So two sides and the corresponding median is b d So a b and then it is they are proportional to the similar setup in another triangle. So hence a b Upon p q is equal to ac upon pr is equal to b d upon q s then the two triangles are similar This we have to prove. Okay, fair enough. So how to prove how to prove These two things now. So clearly a b by p q is given. We have to prove that these two triangles as when when will these two triangles be similar if you have to prove a b by p q is equal to somehow a d by It is already given that a d a sorry ac It's already given that a b by p q a b by p q a b by p q is equal to a c by pr it's given and by uh ac by pr is given. So hence it will also be equal to a d upon ps a d upon ps y because a d is nothing but half of ac and ps is nothing but half of a pr half of pr Okay, so when that is true, that means if you consider these two triangles which which two Yeah, so in in triangle. So let me just write here in triangle in triangle a b d And triangle p q s Okay, what's given a b by p q is already given equal to or we just prove it proved it that it is equal to ad by ps and by given also criteria also this equal to b d b d upon q s It's here So a b by p q is equal to b d by q s and we just proved that is equal to ad by ps also here So that means by s s s These two triangles are similar. Which two? These two triangles are similar When these two triangles are similar my friend then this angle must be equal So hence If these two triangles are equal then what will happen? So Uh, so these two do you use two triangles are equal and the corresponding sides are anyways given to be equal Why if you see a b by p q is equal to b d by pr. I'm sorry b d by Oh, no, you can't take this one ac by pr if you see a b by p q is equal to ac by pr ab By b q is equal to ac by pr and these two angles. We just proved Equals so hence the two triangles must be similar Okay So this is slide number 21 next Let's go to Next slide if two sides and a median by setting the third side of a triangle are a strictly bit proportional now In the previous case, what did we learn in the previous case? It was Now in this was slide number this was slide number 21 So in the previous case In the previous case There were two sides and a median bisecting Median bisecting one of these sides, but in the next slide This is a median bisecting the third third side. So if if you see the next slide The question is if two sides and a median So again if two sides, these are two triangles triangle abc and let's say p q r triangle abc And triangle p q r What does it say it says that if two sides and a median bisecting the third side of a triangle? so this is let me pick ab and P q so let's say these two sides are corresponding sides These two sides are corresponding sides. So ab and P q and ac and pr and the third median of the third side Yep me bisecting the third side. So this is a median. Let's say abc. Let's say this is d and this is s Okay, they are proportional then again the two triangles are similar that what does it mean? It says that if if if ab upon p q is equal to ac If ab upon p q is equal to ac upon ac upon what pr pr Is equal to ad upon p s then then these two triangles triangle abc Similar to triangle p q r This side, no worries. So let's say the ratio of two angle to ratio of areas of two similar triangles is equal to the ratio of The squares of any two corresponding side. So you have already done this multiple number of times So then quickly cap this one and then this one again And then the ratio of areas of two similar triangles is equal to the ratio of Any two corresponding sides. So let's say abc abc and def Okay, ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides Okay, so it's very you know. So, uh, what is that area of triangle abc? By area of triangle def def is equal to nothing but ab square by def square is equal to bc square by def square is equal to ca square by fd square Right again, how to prove that again you have to drop a perpendicular from here Right def let's say bc it is p and this is q Okay, and then this is a perpendicular you dropped Okay, and then you will know that it's very easy triangle abc quickly I'll recap it triangle abc divided by triangle triangle def Into bc into ap upon bc into ap upon half into base into height that is ef into dq Right now half of get cancelled now somehow if I prove that Uh ap by dq ap by dq is equal to let's say Uh, or or or if you prove that somehow ab upon De ab upon de is equal to ap upon dq Then we are done and it is easily you know, uh Easily noticeable how because if you see this angle is equal to this angle and ap and uh, and this is 90 degrees So i'm writing here for the lack of space here. So i'm writing here. So let's say triangle consider triangle abp and triangle de abp and deq Okay, so what ab okay ab Uh, yeah, so a in this in these two triangles angle b is equal to angle c Angle e sorry and what angle apb is equal to angle de q e is equal to 90 degrees Apb is equal to de q is equal to 90 degrees that means these two triangles which two These two triangles these two triangles are These two triangles are similar if these two triangles are similar then what can I say I can say ap by de Is equal to ap ap by de q And if you use this in this so this is what we had to prove And hence this can be written as bc Okay, so it was it was ab by ab by de is equal to ap by de q and since these two triangles were similar So this could be also equal to bc upon Ef because they were similar triangles so hence ab by d ab by de is equal to ap by Dq which is equal to bc by Ef why bc by ef because ab by de was equal to bc by ef because these two triangles were similar So hence from your you can say this is bc square y e square Okay, now clearly Square You can see instead of bc by ef you can write ap by dq here Why because bc by e equal to ap by de q. So this is also equal to ap square by Dq square. It's also done squares of corresponding mediums and We just some if you know previous slides you saw that In if two triangles are similar then they are currently Proportional right in the previous slides So this is also done Also saw this in the previous slides so angle bisectors are also there So if two for two similar triangles angle bisectors are also similar Yeah, sorry also proportional Okay, clear. So this is card number 23. So this is another if you remember we were talking about sorry Yeah, so here is similarity of triangle applications is Lots of we saw the lots of applications and including the ratio of the areas. Okay. Now, let's go to carbon 24 yeah If the areas of two similar triangles are equal then the triangles are congruent that is equal and similar triangles are congruent, okay, so very This also follows all from You know, so they are saying if the areas of two similar triangles. So do two similar triangles are There are two similar triangles a b c a b c and p q r And their area is same So area is same. That means area of triangle a b c Is equal to area of triangle P q r that means That means area of triangle a b c divided by P q r one which is equal to nothing but a b square by a b square by p q square Is equal to b c square by to a square is equal to c a square by here square And all all the ratios are one that means from all this you can say a b is equal to p q And b c is equal to q r And c a is equal to E r ends This side is equal to this side. This side is equal to this side and this side is equal to this side. So So these two triangles are congruent Okay, bye s s s criteria Okay, next Next slide is If a perpendicular is drawn from the vertex of the right angle of angle to the hypotenuse then the triangles on both sides So the perpendicular are similar to the whole triangle And so another this is this we have asked you in multiple occasions also So you this so So there's a right angle triangle first We have dropped a perpendicular from here So that says if a perpendicular is drawn from the vertex of the right angle of a right triangle So this is the right triangle right angle dropped a perpendicular. Let's say this is a this is b This is c and this is d a b d is dropped b d is perpendicular drop down a c, okay Then the triangles on both sides of the perpendicular that that means that they are saying that triangle a d b is similar to triangle a b c a b c and they it's together is similar to so a d b is Similar to a b c and triangle a b c is similar to angle a b c is similar to c d d angle b dc isn't it So hence if you see angle a d b a d b is equal is similar to a b c very easy Why because in these two triangles if you see angle a is common angle a same So what all angles they take we're trying those we are taking we are trying to be taking this this one This one and let me choose another color to highlight it in different colors. So let's say i'm choosing This one these two triangles Okay, so i'm trying just the outline of it. So if you see this is common angle to both This 90 degree is equal to this 90 degree hence by a similarity. They are Similar so angle a is equal to angle a Angle d which is 90 degrees is equal to angle b which is again 90 degrees. So hence Both are similar now with between a b c and c d b same thing happens. So we are saying triangle a b c and triangle and triangle c d c d b these two All right, so hence common angle is c common angle is c both And d is perpendicular angle d this this angle is 90 degree and this angle is 90 degrees So hence b is equal to d right perpendicular both are perpendicular. So hence again, this is Similar so hence they are similar to each other If you remember I had given you this question sometime back In your module desks, okay Fair enough. So let's move ahead. We are running short of time. We have four slides small to go So we'll complete. Okay Pythagoras theorem in a right triangle the square of target news This you have this is so many times you would have done So let me not go again. So if you want again This is what the proof looks like abc theorem says ac square is equal to ab square plus bc square you have to use what first of all you have to use triangle a d b is similar to triangle a b c And then using this you have to find out one one relationship between ad's Let's say between those two sides that is the corresponding sides and then using other other two sides that is angle triangle c d b is similar to triangle c b a using these two triangles these two similarity you can Prove that I am not going in the details again Yeah So if triangle one is similar to triangle two and triangle three is also similar to triangle two Can we just say that triangle way and triangle three are similar or do we have to prove it separately? No, you don't need to You don't need to you don't need to prove it separately because it's it's clear Evident from let me do it once again. So let's say I'm taking this so triangle clearly. This is done When adb is similar to abc and cbd is you know, but you have to just maintain the order of The vertices so let's say adb is similar to triangle abc is similar to Triangle what c remains at c's position b goes to d and becomes bdc. Yeah, so And this is how it is Okay, so you can you can do that directly you don't need to prove them again Okay, next so this is also done. So pythagoras theorem. I am again leaving it on to you to do the The next slide Converse of power So converse of pythagoras theorem is again by contradiction. You can again use so what is converse of pythagoras theorem? It says It says that in a triangle In a triangle abc if ac square is equal to ab square plus bc square then then ab Angle abc is right angle So let's say a b ab ab ab c is the abc is the is the triangle where ab ac square is equal to ab square plus bc square Okay, and then We have to do what to prove that angle b is equal to angle b is equal to 90 degrees So what you do is you draw another triangle. So this is like your ncr to prove maybe def This is def you draw such that what is that d is equal to ab. So ab is equal to de and e f E f is equal to bc or bc is equal to def And angle e is equal to 90 degrees. So you make it like that. You make our triangle Equal to 90 degrees now Clearly if in this triangle, this is construction construction So if you do that, you will get what def square is equal to de square plus def square adharas Right and def square hence can be written as Plus bc And So hence what I'm using this you can write but ab square plus bc square was ac square. So the so Square b square plus bc square it's given from here is ac square. So write ac square So hence def is equal to ac So hence in in both these triangles In both these triangles if you see ab is equal to de and bc is equal to e f and ac is equal to def hence angle Is triangle e e f hence Hence Is equal to Which is This is a conventional proof. So if at all there is this proof you can Use this method. So hence construct another triangle with corresponding these equals These equals and then ac square is equal to you know, you can do that Uh prove the two triangles to be congruent and then then prove that this is equal to 90 degrees This is how the conventional proof goes. I will again, you know, we in the class I we attempted the another Uh proving method will again solve and send it back to you. I'll just read out these theorems so that you are This thing is complete. So hence in any triangle the sum equal to twice the square of half this is Uh, we have done in the class as well. So it says if this is a triangle abc This is actually called apollonius theorem. So any triangles of the squares of any two sides. So ab square plus ac square Is Is equal to the twice the square of half the third side and the median which so hence it is Is equal to twice the square of Twice the square of half the third side. So this is d So two times dc square plus ad square. This is apollonius theorem. We have discussed the proof in our Classes right this is apollonius theorem again. You have to Uh use pythagoras theorem too. So what you need to do is you need to drop up a pendicular here Drop up a pendicular amc. Let's say put it p and then use pythagoras theorem And you know express ab square in terms of ap and bp express ac square in terms of ap and pc and then And then express so there are three pythagoras theorems. You'll write you'll write ac square equations You'll write ac square is equal to ap square plus pc square and then you'll also write ad square is equal to ap square plus pd square and then you'll write ab square is equal to ap square plus bp square and using these three you eliminate ap And you know another another equation is bp is equal to bc is equal to half of bc using all this you eliminate ap and every point related to p And you will get this Okay, so this was abollonius theorem proof now next We are almost done. So let's go to card number 29 Three times the sum of the square of the sides of a triangle is equal to four times the sum of the squares of the medians of the triangle Yes, this is again a very important theorem. You need to Know this but then you you avoid using them directly in your problem solving Actually, they will there will not be any direct application. But yeah, so what is this theorem? This theorem says that Okay, this is abc abc And it says three times the sum of the squares three times ab square plus bc square ca square Is equal to let's say these are the medians and this is the direct application of the previous term That is the apollonius theorem, right? So let's say this is def Is equal to four times d bd square b d square plus af square plus ec square okay af square Three times ab square plus bc square plus ca square is equal to four times bd square plus af square plus ec square How to again use this apollonius theorem apollonius theorem says what ab square plus ac square is equal to twice of af square plus a First of af square plus bf square and similarly times and Then eliminate so ab square and then you write bc square plus ca square is equal to two times bc ca is two times c e square Plus b e square And then again you write third one. This is this is two and then third word. I am writing here and that is nothing but ca square plus ab square ca square plus, sorry, uh, not bc plus ac b road and bc and ca i wrote then i have to write ab square plus bc square, sorry, this is bc square So bc square plus ab square bc square plus ab square is nothing but twice of bd square Plus bc square Using these three using three add all of them together You will get this relation this relation by Saying that and you know that how do you replace dc? So dc is nothing but cd is nothing but half of ac you write that And bf is nothing but or bf is equal to wherever bf is there is bf so bf can be written as half of bc and uh And b b is equal half ab so using these three and these three also together you'll get this relationship You'll have to just eliminate Fair enough. So this is card number 29. So this is also done And uh, what next? I left with this one three times the square of any side of an equilateral triangle Is equal to four times the square of the altitude. This is again fall out of the previous one So hence if you see this is nothing but if there is an equilateral triangle Equilateral triangle So if there is an equilateral triangle, this is equal triangle and they are saying that Three times the square of any side of an equilateral triangle. So let's say abc Right. So three times ab square Is equal to four times the square of the altitude Four times the sign or ab And you can see it's direct direct proportional direct fallout of the previous one. Why the previous one was this three times Uh, what ab square Three times the square of any side of an equilateral. Yeah, so ab square plus bc square plus ab square plus bc square plus what? ca square is equal to four times the median and in this equilateral triangle case median is equal to altitude So four ad square plus let's say ad be This is be Be square plus cf Cf square right now In in case of equilateral triangle ab is equal to bc is equal to ca Right. All sides are equal and altitudes and medians are also equal. So you can write ad is equal to be Is equal to cf Using these three. What can you say from here? So let's like this color Yes, so if from here, okay If from here It is nothing but three square Is equal to four times Ad square will be equal to b square equals to cf square from here So hence three three gets cancelled. So hence three ab square equals four ad square Right. So this is your summary of the entire This, you know, now what remains is problem solving. Obviously these are the You know points to you know, uh Keep in your mind while you are you are trying to solve problems. So beyond these no theorem and no You know application will be there. All the questions you will be getting you will be having uh these only Uh theorems, you know useful now always keep in mind. I told you what did I tell you That just a minute. I'll show you Yes, so this is the basic information Yeah, basic information says that the moment you sequence little triangle What could come to your mind first? use of similarity or In all One or max couple of, you know things to prove related to this thing Uh theorem to be related stuff Otherwise mostly that if there will be two there will not be a direct theorem. There will be application of these So Moment you see a triangle Basic proportionality theorem Similarity of triangles all the additional patterns here No corresponding squares of size and all that Will be included. So once you have the list of all the theorem in front of you Then you know, uh, you will be able to Think through the problem and Solve it. So again, there will be if the problem related is related to ratios Or if you see lots of parallel lines in the diagram or if you're in the form of square, so you The moment it is in the form of square, then you know, either the application of Pythagoras theorem Or it will be the ratio of areas Ratio of areas These are the only two techniques through which this can be solved Okay, guys, so if you have any queries queries questions you can raise, uh, you know um reach out to me and I mean that we are we are you know, recording a lot of problem-solving videos Which are in progress if you want any particular problem to be solved and and Send back to you You can always, uh, you know reach out to me or any of any of the other faculty members Or you can also, you know, uh Mention in the comment section also no problems Uh You know, I actually got a little bit prolonged. No worries. We'll try to Canine to do our session from But I hope it was useful. So next class I believe is on Friday and Has been circulated to all of you. Please try to Attend the session so that it becomes a quick revision of all that is there at least it would leave you an overview of the entire uh All the switch is there and it will be there in your There in your mind okay, so Yeah, so let's call it today and then Uh, we'll meet again. Thanks for your time and keep watching the problem solving videos and Keep solving worships. We have shared a lot of content material that drive as well. So please utilize them try to solve As possible. This is the only way you can Improve your score now Okay, thanks a lot. So we will close Your thanks for your time