 Okay, good evening everyone. I'm waiting for others to join. There are at least eight nine more people to join out here So Let's wait for them. Okay Okay till the time they join I'll give you a question you solve it so till now we know Tales theorem what we already know I am writing it on the so We have already done tales theorem and We have also done Criteria for similarity in that we have completed triple angle similarity Which leads to double angle similarity then We have side side side Similarity and then we proved in the class last time side angle side similarity So now we would be solving question based on all these things So, that's how we also solved Perpendicular theorem so I'm not Adding that now The question is There are we will solve three more theorems. So you will solve it the Additional theorems are if two triangles are equi-angular and equi-angular means Angles are same corresponding angles are same prove that the ratio of their corresponding sides It's same as ratio of the corresponding at altitudes So I'm drawing figures for you guys in this I'm sorry. So, suppose I have a triangle like this. I Again have a triangle like this. I Draw an altitude like this. I Again draw an altitude like this. I am naming it This I'm naming it as a B C L and This I am named naming it as D E F and this as M So you have to prove that by DE is equal to BC by EF is Equal to I'm assuming that angle A is equal to angle D because it is equi-angular So this is given to you an angle B is equal to angle E angle C is equal to angle F This is given to you So you have to prove that AB by DE is equal to BC by EF and that is equal to AC by DF and That is also equal to AL by DM So you don't have to prove this actually because of with A Similarity you can write this you just have to prove that any one of them is equal to AL by DM That's what you have to do Solve it quickly. I give you two minutes those who are joining now a Question has been written on the screen. You have to solve it figures. I have already drawn good Your method is right, Vanessa. Okay. Should I solve it? Screen quality bad for all of you Hanish Okay Fine. So look at here a BC I'm writing solution out here if ABC and DEF are Equi-angular Then I can write that ABC is similar to DEF By angle angle angle similarity so if this is the case then AB by DE is Equal to BC by EF as it has been written How will I prove that this also comes into picture to take this into picture? The concept is how you should think about this question that is very important. So when when I draw these two When I draw these two perpendicular, so this perpendicular divides Triangle ABC into two parts. I'm not saying equal parts. It may look like equal. It may not be equal. So for Your clarification, I'm not saying equal parts Similarly, this DM is dividing this triangle DEF into two equal parts Now if I prove that one part of this triangle is Similar to one part of this triangle, then I'll be able to prove that AB by EF would be equal to AL by LM So the concept that I should apply here or the thought process should be that I prove one part of this triangle So for this for for this question, let's say I am proving ALB I have to prove this if I am able to prove this ALB is similar to D ME Then I'll be able to write AB by DE is equal to AL by DM so My question would be solved now. How will I prove this? So try to understand you look at here Angle L is equal to angle M. So why because it is an altitude. So this is 90 degree So this has been given to me and as it is equi-angular So angle B would be equal to angle E. So this is given in the question Now we all understand that Already by double angle similarity, I can write that ALB is Similar to angle DME and when I am able to write it, I can write that AB by DE is equal to AL by DM and I can also write corresponding side. So Question may be turned that the ratio of the sides on which altitudes are drawn Altitudes have been drawn on side BC and EF So I can similarly write BC and EF because AB by DE is equal to BC by EF. So this is how it is So I hope you understood it The second question is for those screen is not clear you increase the quality So Screen quality can be adjusted from the YouTube instead of 360 you make it 480 or above Screen quality would be good in that that case Similarly second theorem of today's class is if triangles are equi-angular prove that the ratio of their Corresponding sides as the ratio of their corresponding medians solve it. I'm not drawing Once you are done. Let me know Then I'll start solving because I don't like to wait much so If you're done, let me know. Okay. I'm solving it now. So Again first I'll draw the figure for you and The figure would be something like this. This is one figure and This is the another one I draw a median for you. So This is my median and I name it ABC again. I name it ABC This is L and this is DEF And this is M if it is equi-angular so angle A is equal to angle D angle B is equal to angle E and Angle C is equal to angle F So this has been given and if this has been given so I can write that triangle ABC is similar to triangle DEF and And in that case I can write that AB by DE is equal to BC by EF is equal to AC by DF is equal to AL by DM So this I have to prove Proof now again the thought process should be if I have to include this median then I'll have to prove that So I'll have to prove that That the triangles formed by these medians inside the original triangles are similar so I am taking ABL is similar to D sorry It's similar to DM E and To prove it. I already know that angle B is equal to angle E. So there is no problem out there now Try to understand So look at here if it is a median. So if it is a median what happens is AB by DE I can already write here and what is getting divided into two parts BC and EF is getting divided into two parts. So I can write this BC as so BC equal to to BL and EF here would be equal to 2 EM So I can write BL by EM So side side similarity is there and if it is median. So what happens is so as I have taught you last class SAS similarity in SAS similarity what happens that Ratio of the two corresponding sides are equal So AB by DE is equal to BL by EM and the angle in Angle made by these two corresponding sides are also equal. So I can write here Angle B is equal to angle E and if this case is in front of me I can write that by SAS similarity ABC is similar to DEF So in that case I can write that AB by DE which is also equal to because corresponding sides have been asked BC by EF That is equal to AL Divided by DM. That is how you solve it Now next one is same thing I'm not writing the question. I have written two questions the may replace median Because it's taking a lot of time in we have done it for perpendicular altitude. We have done it for medians replace altitude or Median with angular bisector So take angular bisectors of angle A and angle D correspondingly and then prove it And this is very simple Angle is the included one vocal if you go to the previous question. Let me take you to the previous question Angle is included one. I'm saying AB by D. So I'm sticking this side I'm saying BL by EM. So I'm taking this side Angle B is equal to angle E. So angle B is included in AB and BL Angle E is included in DE and EM. So this is included angle Pradyat, you cannot apply a similarity because there there is no concept of a in in case of medians You have to apply SAS Once you are done, let me know. I don't think that can be solved by a because there is No possibility of a there Okay, so a lot of people have done it. So let me take This is one So this is a BC and this is DEF again the same methodology So angle I'm just solving it quickly angle B is equal to angle E and I know that angle A is equal to angle D Now AL and DM are angular bisectors. So hence they will be bisecting angle A and D So half of A is equal to half of D So I can write that by angle angle similarity ABL is similar to DEM So by that I can write AB by DE is equal to BC by EF and that is equal to AL by DM So these three theorems are proved now, let's move to questions so Questions are and this is CBAC 2002 question so Approve that for similar triangles the ratio of their corresponding sides is equal to ratio of their Perimeters quickly in this In this question giving hint when solving the question itself so wait for two minutes Good two more minutes nothing more than that. Okay. Now look at here. So ABC And DF So if two triangles are similar I can write that and an opposite side. I'm writing this as a this as B and this as C And this as D this as E and this as small F So a by D Is equal to B by E is equal to C by F So this is already given I have to prove that that this is equal to a plus B plus C Divided by D plus E plus F So now suppose I prove it I write it equal to K so a is DK B is equal to E K and C is equal to F K Put it here. So this gives me DK Plus E K plus F K Divided by D plus E plus F take K common D plus E plus F Divided by D plus E plus F So that comes out to be equal to K. It means that if this is equal to K. This is also equal to K So that's how we prove this so we can write that This is equal to perimeter of triangle ABC divided by perimeter of triangle DF Well, that's how you prove it Okay, next question This is a triangle and I draw a perpendicular out here. This is one perpendicular And This one is another perpendicular So I name them a B C D E and this is perpendicular. This is also perpendicular So they intersect these two altitudes intersect at point P You have to show that or prove that first one is triangle AEP It's similar to triangle CDP CDP second one is a triangle ABD is similar to triangle CBE third one is triangle AEP It's similar to triangle ADB and fourth one is triangle PDC It's similar to triangle BC This is P If you are done, let me know So I've not got one answer Yes, you can with them. I've got only one answer as done, okay You take two more minutes Okay, so a lot of you are giving answer that you have done it One person is saying that Wait a minute. I'll wait for one more minute for him Okay so now let's start solving these questions and I'm doing the first part of it. So Have a look at it for first part tell me to prove that that triangle AEP Has to be proved similar to triangle CDP So let's go to triangle AP if you want you can redraw it So this one is triangle AP and this one is triangle CDP So I know that here angle E is equal to angle D because they are both 90 degrees So it has been given to me now This angle is equal to this angle. So I am writing angle APE is equal to angle C PD why because they are vertically opposite angles, so they would be equal So I prove by angle angle similarity that these two triangles are similar. I don't need to prove much second one is Triangle ABD is Is similar to triangle CBE so So try to understand ABD where is ABD ABD is this triangle this particular triangle ABD and CBE is CBE is this triangle So I can write angle B is equal to angle B because angle B is common between both the triangles and angle D Is equal to angle E because they are 90 degrees. So this is angle angle similarity similarly, the third one is angle a triangle AP so AEP is again this triangle and triangle ADB So again triangle ADP is this triangle now try to understand here also angle B is equal to no, sorry A triangle AEP so AEP So once again, just give me a moment. This is not right. So I'm just erasing it So and I can write angle E is equal to angle D that is 90 degree and Look at here Okay, so this angle is common between both the triangles it is common in AP also and it is common in ABD also So I can write angle BAD is equal to angle EAP because angle A is common there. So again, I can Prove AA similarity and let me go to the fourth one. So PDC Let me solve fourth one here so that you can understand it. The figure is also not erased So fourth one is PDC and BAC. So where is PDC? PDC is this one and where is BEC? So BEC is this bigger triangle now try to understand in this PEC and BEC this angle is common. So I can write PED Sorry PCD Let me erase it So that it doesn't create any confusion So PCD is equal to angle ECB angle C is common And I know that angle D is equal to angle E. So again AA similarity. So that is also proved So they're very very simple question Now let me go to A question um And there is no figure in this question. So I love this kind of figures. I don't have to draw one So question is In a triangle ABC Angle sorry side AB is equal to AC And D is a Point On AC Such that BC square is equal to AC Into DC prove that BD is equal to BC quickly Do it. Yeah Charan My power is low. I just quickly put into charge Once you're done, let me know Do it quickly Okay, one answer Two people have done I'm waiting Okay, I've got a lot of answers till now. So I mean I've got a lot of people saying that They've done it. So Let me solve it So let me draw a triangle first and the triangle is like this and then Let me draw a line first and the line is something like this Okay, and Let me name it then. So this is a this is B. This is C and this is D now It has been given two things have been given So I am writing things given to me given is AV is equal to AC I know angle B would be equal to angle C because It is an isocellous triangle similarly I have been given that BC square is equal to AC into DC So this is also given to me. So these are the given things now this BC square by BC square is equal to AC into DC can be written as BC by DC Is equal to AC by BC Why am I writing this because if I have to prove this BD is equal to BC Somehow I have to utilize the formula of corresponding angles Corresponding sides that the ratio of corresponding sides are equal And to do so I have to prove similarity and that is why this particular equation I am writing it in terms of the ratio of corresponding sides. So I know that I am writing BC By DC is equal to AC by BC. So I It means that if I take this and I write like this BC by AC Is equal to DC By BC that will also be a Way to write it and I know that if I am talking about BC DC AC BC then it means that I am talking about two triangles and these triangles are triangle ABC and triangle BDC So in triangle BDC, obviously BC by AC is equal to or BC by DC is equal to these From these two things ratio of corresponding sides have been taken and I know that Or let me write like this. This is BC by AC Is equal to BC by BC. I'm talking about only these two sides Hence the included angle is angle C. So I write angle C is equal to C And this proves by SAS similarity that triangle ABC is similar to triangle BDC And if this is there, then I can write that in in in the same similarity. I can write that AC divided by BC Is equal to AB divided by BD Now I know that AC is equal to AB So this and this gets cancelled out. So I can write BC is equal to BD. So that is how you prove it Next question Okay, next question is Again, no diagrams. So in Triangle ABC P and QR midpoints of AB and AC respectively Such that PQ is Parallel to BC Prove that median drawn BC from A And median name is AD median AD drawn to BC from A bisects PQ Quickly done No one has done yet Okay, wait two three moments Okay, so Three people have done it till now. Okay I will wait Okay So a lot of people have done it now. So, um, I can write the answer I can I can Start giving you the answer of this question And this question tells me that so let me make a diagram for you and this looks something like this And I then I draw a line like this So and then I know how to mark it So this is a this is B. This is C And these are the midpoints. So this is P and Q And I have to draw one more line. So Let me take Different. So this is nothing but D So, uh What has been given to me this has been given to me that AD is a median So BD is equal to DC. I know AP is equal to PB and AQ is equal to QC because they are midpoints And now what I can do over here is that I know that Try to understand if I have to prove that these I have to prove that suppose this point is E So I have to prove that let me write with a different color So to prove that PE is equal to EQ If I have to prove that PE is equal to EQ Then somehow or the other I have to apply this formula of Similarity of triangles where I will be writing a ratio of corresponding sides and then I'll prove it So to do this what I can do is Let me Take triangle APE And triangle ADB So I know that this angle is common. So angle PE is equal to Angle BAD This is given Angle APE So this and this would be equal. Why because these two sides are parallel. So these would be corresponding angles So angle P is equal to angle B because they are corresponding angles So corresponding angles So two angles are same. So I can write by a similarity that triangle APE is Similar to triangle AD B So I can write that AE by AD Is equal to PE by BD and I also know that BD is equal to DC Now what I'll do I'll go to the other side of the triangle So I know that I don't I am not proving it You can prove that this angle is common and this angle and this angle are equal. So you can write that triangle A QE is similar to triangle ADC And you can write here that AE by AD Would be equal to EQ Divided by DC Now AE by AD is common both the sides. So from if this is equation one and this is equation two from these two equations What I can write over here is that PE by BD Would be equal to EQ by DC Now I have already written here that BD and DC are equal because this is a median out here So this can be cancelled out and I can write PE is equal to EQ Which tells me that E is the midpoint of this line PQ or E is or this line AD is or AE is bisecting PQ So I hope you understood it Next question is line segments Joining the midpoints of the sides of a triangle form four triangles each of which is similar to the original triangle Solve it quickly You can do it Mehta You can do it. You can make parallelogram out there Ankit try Ankit Raj Ankit Okay, how many of you have done it? Yes Okay Charan has done Raj Ankit has done A lot of people have done I have not seen this, huh? Okay, let me solve it. Let me go out here and let me draw it for you So it's something like this Then I'll have to draw lines So I draw a line like this I draw a line like this Oh my god, I draw a line like this And then I draw a line like this So I name it A BC And this I name it as DEF So I have to prove that So what are these? These are the midpoints of the sides. So DEF are midpoints of BC, AC and AB respectively So I have to prove this that triangle AFE This is one triangle Is similar to ABC Triangle FBD It's similar to ABC triangle EDC Is triangle similar to ABC? And triangle DEF It's similar to triangle ABC Now, how do I prove it? So let me prove it for you So let's say that FNER midpoints of A and AB and AC respectively So F and ER midpoints AB and AC respectively So apply midpoint theorem Or write by Thales theorem Which is our converse of Thales theorem Which is nothing but midpoint theorem proof of midpoint theorem FE would be parallel to BC So hence F and BR equal A is common A is equal to A F is equal to B and C is equal to E. So hence proved Similarly, this this is So everything is same. You just have to show that this is Parallel to this and this angle and this angle would be equal this angle and this angle would be equal This is common. So you can prove that BFD is similar to ABC. Similarly This and this are parallel by midpoint theorem. So you can show that DCE is Similarly for EDC Similar to ABC now You only need to prove that DEF is So if to prove this DEF I have to prove this only now So let me go here and let me write That now we need to show that triangle DEF is similar to triangle ABC and to Prove it what I do write is that Look at here ED is Parallel to AF And what is AF? I'll show you here. I'm saying that ED is parallel to AF Why I'm writing parallel to AF because ED is parallel to AB and AF is part of AB And that is why I have been writing ED is parallel to AAF I can also write that FD is Sorry FD is parallel to AE So opposite sides are parallel. So I can write that AF DE is a parallelogram It's a parallelogram Now if it is a parallelogram then Opposite opposite angles would be equal. So this angle Would be so angle D is equal to here angle D. This angle D is equal to angle A And I can write that Similarly, you can show that show that Uh, I'm making a parallel this as parallelogram again. I can make this as parallelogram this one So I can show that BDEF is also a parallelogram and from there angle B would be equal to angle E So this is equal to A. This is equal to B. Similarly, I can show that F is equal to C So all three angles are equal. So I can write that B is equal to E And by from here only I write angle angle similarity and it will be proved So no worries out here Okay, wonderful. So let's have a question now on Obviously you can use but congruency. How will you use? I don't think you can use congruency or size sides have never been given here So Try to understand. Let me draw a figure for you. This is Like this It's something like this So it's something like this. So let me now write its name And let me give you the properties. So this is a This is B. This is C and this is DE This is FG and this is 90 degrees So you have to prove that And it has been given that DE FG is a square Given angle A is equal to 90 degrees Given proof triangle AGF Is similar to triangle DBG Triangle AGF Is similar to triangle EFC Triangle DBG it's similar to triangle EFC And fourth one is DE square Is equal to BD into EC This is first This is second Third, this is fourth Hey focus on solving question not teasing each other Give me the answer Good Why people have left only 10 people are here Who all are there? Just let me know. Let me check how many of them are not Pat And don't message them that sir is asking for attendance I've got attendance of at least 12 people and and and it is showing only nine people are watching it. How come? Okay, maybe Chalo, let's solve Solve give me the answer only one answer I've gotten now Okay, fourth one. You are not getting chalo. Let me solve So let me first go to AGF. Where is where where is AGF? This is AGF This one is AGF and DBG So DBG is this one. So I have to prove similarity of this and similarity of this How will I prove it? So this angle is 90 because it is an square and this is 90 So for first one Okay, let me write Let me change the color first. Okay for triangle AZF And triangle dvg Angle a is equal to angle d No doubt about it And if this is a square so this angle is parallel to this angle Sorry, this side is I can write that de is parallel to gf why because square sides So this angle will be equal to this angle. So angle b is equal to angle g Because they are correspond corresponding angles. So corresponding angles So by a similarity, I proved that these two are similar. So first one is proved Second one is AGF and EFC. So similarly, this is 90. This is 90 and this and this are similar. So that is proved DBG DBG and EFC. How do you prove it? So this is 90 and this is 90 So third one second one can be proved. Similarly like first one third one is triangle dvg And triangle EFC How do you prove it? So you can write that or you can write that angle d is equal to angle e and b is equal to g and c is equal to f So try to understand or Better solution would be that this triangle is similar to this triangle. This triangle is similar to this triangle So if this triangle is similar to this triangle, this triangle is similar to this triangle So these two triangles has to be similar. So from one and two only this is getting proved. No need to prove it separately So let me go to the fourth one. The fourth one is what I can write triangle dvg is similar to triangle azf and triangle EFC is similar to triangle azf From here you can prove this fourth one Where were you Karthik? After two minutes three minutes you are saying you are here Left the class in between when to enjoy somewhere, huh? now as triangle Look at here bd is here ec is here and dc is d is in between So bd and ec are part of triangle dvg And I know that triangle dvg is similar to triangle ecf So I can write that bd divided by ef or bd divided by dg Is equal to dg ef divided by ec now I can write that bd by ef Is equal to dg By ec So I can also write that One second Now look at here this Where is dg one second? Okay, I got it So dg and ef are equal because try to understand look at here. I'm writing one thing that d e is equal to ef Is equal to fg is equal to dg Why because they are sides of a square So I need bd. I need ec. So you look at here bd into ec from here is equal to dg into ef Dg and ef both are sides of square So dg and ef both are sides of square They can be replaced with de because that is also side of square So that can be written as de square Is equal to bd into ec Okay, so this question is done Let's solve next question This question is like this How many more till the time 8 o'clock? I am not bored. You are bored already, huh? or through the midpoint m of the side cd of a Parallelogram abcd the line bm is drawn intersecting ac in l and ad produced to e Prove that el is equal to I cannot draw the diagram. I don't know it Poppers fully. I have given this question so don't kill the poppers Of giving this question to draw it yourself The sole purpose of giving this question was to make you draw the diagram yourself Hey, it's all the question Half the time you are like bro, bro, bro, huh? Okay Fine one person has done it Okay, so this is the diagram Now in this diagram, I have to prove that And this is l they are intersecting in l. So I have to prove that al is equal to 2 bl So how do I prove it? So look at here Let's stick different angles. So let me assume angle. Suppose this this is angle one And this is angle two This is angle three This is angle four This is angle five This is angle six So I take triangle triangle bmc And triangle emd So I take these triangles. So I know that Angle one is equal to angle two because they are vertically opposite angles I know that mc is equal to md because m is Midpoint angle bcm So this angle is Try to understand This angle would be equal to angle this angle Now you ask me. Um, no, not this angle. Uh, this particular angle Why this angle would be equal to this angle try to understand angle b Cm is equal to angle edm because If they are parallelogram, so this side and this side is parallel and this is so I am writing here bc is parallel to ad and dc is dc is Transversal So if dc is transversal, then in that case what will happen angle d will be equal to angle c So whatever I written this is nothing but congruency so triangle bmc Is congruent to triangle edm So if this is the case then by cpct I can write by cpct bc is equal to de this x side and If bc is equal to dc de I know that bc is also equal to ad Why because they are opposite sides of parallelogram. So I can write that ad is equal to de So I can write bc is equal to ad that is equal to dc So you look at here then it can be written as if ad is equal to de So ae Is equal to ad Plus de So that can be written as to ad or it can be written as to ae because they are all equal Or it can be written as to bc. So anything is possible out here Now, let me go to another triangle. I have to prove that bl and en So then I have to take triangle bl is part of triangle cbl and El is part of triangle ael So triangle ael if you are not able to Look at the diagram try to understand bl is part of this triangle cbl And this is ael. So what I'll write over here is I'll write that Angle 5 is equal to 6 which is vertically opposite angle now angle 3 is equal to angle 4 And which one is 3 and which one is 4? I'll show you now Now look at here this angle and this angle would be equal because they are Alternate angles. These are the two parallel sides. They are alternate angles So I can write it as alternate angles And if they are alternate angles then by AA similarity, I can write that triangle cbl is Similar to triangle ael if they are similar then I can write that el By bl Is equal to ae This el by bl is equal to ae Divided by bc Now, I know that ae by bc is equal to 2 From here ae by bc is 2. So el by bl is also 2. So I can write el is equal to 2 bl So this is what the question is now your homework is After this topic I've done most of the examples after this topic Whatever exercise comes so there are four exercises in this question in this Chapter triangles the first one is after the el's theorem. The second one is after similarity this particular topic which we are doing so after similarity whichever Exercise is there. You solve all the questions of that exercise before next friday. I'll be checking that plus After this You have theorems of Ratio of areas of two similar triangles. You solve most of the Salt examples from there So that's your homework Charon you cannot use congruency here. There is no Chance of using congruency. Congruency can only be used when when the sides are equal Here there is no relationship between Sides have been given so r s agarwal from r s agarwal solve exercise after similarity of triangles plus after that exercise Theorums have been proved about ratio of area of triangles for similar triangles. You do that Do the salt examples and then let's meet on next friday. Thank you so much for joining the class Bye