 Hello and welcome to another session on similar triangles in the previous session guys. We Show showed to you how a a a similarity criteria works Correct. So what was a similarity criteria? So let's first write down the statement and then we'll try to establish it through a general prove Okay, so the statement suggests that if if to triangles Triangles have Equal corresponding angles Okay, that means corresponding angles are equal in it in two triangles Corresponding angles then the two Triangles are similar Triangles are similar, right? This is the beauty of Triangles as such Because you don't need to prove or establish both the criteria of similarity, right? So only one will anyways ensure the other one so what are the two criteria one is the two the two triangles must have corresponding angles equal and The two triangles must have their ratios of corresponding sides also equal So that means it's already given that the angles corresponding angles are equal That means we just need to establish the fact that if the corresponding angles are equal in a triangle That will automatically ensure that the corresponding sides ratios are also same Then we are done So let's try and draw a triangle first Okay, so here is our triangle One triangle right and let's try to draw a triangle similar to this Okay, so what I'm going to do is I'm just copying in Creating another copy of it. Okay, here it is but little bigger in size. Okay, so Appearance wise it does look like That both of them are similar. So let me write the names first so ABC and The other one let's say D E F Okay These are the two triangles and what is given so let's write down given as well. So given is Let me use yeah, so given is angle a is equal to angle D Angle B is equal to angle E. So I'm not writing the full names. You know, what does it mean? Angle C is equal to angle F this is given. So what do we need to prove? We need to prove what do we need to prove is a b by D e a b by D e is equal to BC by E F and Is equal to be sorry CA By F D This is what we need to prove. Okay, so the given angles are equal. So let's Mention that so this angle is equal to this angle D This angle. So let me use another color so that it becomes easier to identify. So this angle B is equal to angle E and Angle C is equal to angle F Okay, clear, we have to prove AB is equal to D. E. Sorry AB by DE is equal to BC by E F is equal to CA by F D Now, how do we go about it? So we need to do some constructions obviously the tools available with us is that of congruent triangles angles some property exterior angles And all that what we have already learned. So we are going to use And obviously the basic proportionality theorem, which we have learned Just some sessions before. Okay, so what I'm going to do is this that Now if there are two triangles with not equal sides, then one has to be equal One has to be smaller than the other. So there are three cases. What are the three cases possible case one is That The ratio is equal to one. So let me say this is K. So this ratio is K. AB by DE is equal to BC by EF Is equal to CA by F D and is equal to let's say K is equal to one. This is the first case K is equal to one K is equal to one means AB is equal to DE Correct, this will be happening only when AB is equal to DE BC is equal to EF and CA is equal to FD now I'm saying let us say that this holds. Let us say this holds We are taking up. We are taking two triangles where corresponding sites are equal then Then we don't need to prove anything the ratio is anyways one right in that case in that case So I'm writing if this happens. So let me just take away this K is equal to one It will unnecessarily confuse you. So if if this condition is fulfilled the first one, which I'm mentioning AB is equal to DE BC is equal to EF CA is equal to FD then we then we can say What can we say that then then? Then AB by DE Is equal to BC by EF Is equal to CA by FD proved in this case it is proved and all are equal to one hence proved But this is only one condition guys only one you know one case where all the sites are equal where where in AB was equal to DE BC was equal to EF and AC was equal to DE F right, but that need not be the case every time So hence the other case the two cases will be other two cases are that K is less than one or K is greater than one or let's say AB is less than DE This is one case and third case will be AB is greater than DE Now case two and three are all same. Why because it's just a matter of naming right whether you name AB C To the left triangle and right triangle you name is DEF or the other way around Right, so I can I could have called this as DEF And this one is ABC hardly matters right so nomenclature or naming a triangle would not Alter the result that they are similar Okay, so hence what I'm saying is If that is the case that you know, uh, let's say we are saying either of these case So what I'm trying to hint upon is whether you start from this or start from this The proof remains the same for both the cases because you can just swap the names of the triangles and Get the same result. So I'm saying let's say I'm going to prove For AB less than DE if this is so then what happens? Let me take this drawing Where we are proving it so here is the drawing I'm copying and pasting So I brought the triangles here guys now what to do. So what I'm doing a construction now So what is the construction construction is I mark a point Here E dash And here a point F dash Okay on DE and DF respectively construction. So I'm saying A E dash is equal to sorry not a E dash DE dash I'm I'm constructing constructing DE dash DE dash equal to AB and D F dash Is equal to AC I can do that because AB and AC AB is less than DE so all all are less than You know corresponding sides are less than the corresponding sides of the other triangle, right? So I can do that and what do I do is I am joining E dash F dash joined E dash F dash Okay, joined E dash F dash now consider in triangle Consider these two new triangles. So ABC is one And this another triangle a dash. Sorry not a dash DE dash F dash Okay, let me write it clearly No confusion should be there. So DE dash F dash. What can I say? AB is equal to DE dash by construction By construction, isn't it Then angle A is equal to angle D. It was given the three angles are Correspondingly equal and we know that AC is equal to DF dash by construction again If these is true, then what do we get? Therefore By SAS criteria, I can say triangle ABC is congruent to triangle DE dash F dash Correct by which criteria by SAS S congruence criteria Clear no problem in this. So this will lead to a lot of many other interesting results. What are they? So the results are something like this. So this angle is let's say X Then this angle is X because CPCT And this angle is Y So this angle is Y Right, so this is what we are going to we are going to get so many other Important results as well Okay, now If this is so then how do we get to the desired result of equal ratios now guys It is We can say that angle B is equal to angle DE dash F dash Y congruent parts of congruent triangles CPCT And angle C is equal to angle B F dash E dash Y it is same CPCT But angle B was equal to angle E given And angle C was equal to angle F. This was also given So from these four What can we infer we can say that? So let me write it here. So angle I'm writing here angle DE dash F dash is equal to angle E Right, that means I can write this as X The moment that is done. I don't need to go for any other thing. We can say this say then They are corresponding angles, isn't it? That means E dash F dash becomes parallel to EF E dash F dash becomes parallel to EF. Why? Because angle DE dash F dash and angle E Are corresponding angles which are equal We just proved corresponding corresponding Angles we just proved this Okay, just prove this Correct. If they are corresponding angles, that means This is established parallel and now in triangle in triangle DEF We have E dash F dash parallel to EF. Does it ring the bell? That is we can use the BPT, isn't it? Therefore by basic proportionality theorem, what can I say? I can say DE dash By DE or the corollary to BPT I can write this Is it it? But DE dash was what? If you check here DE dash Was AB and DF dash was AC. So can I not come back and write this as AB by DE Is equal to AB by DE is equal to BC Or the other way now wait a minute Yeah, is equal to BC by BC not BC DF dash, right? So AB by DE is equal to Is equal to AC DF dash is equal to AC by DF So we establish one of the Rations which was supposed to be done. See what was to be proven So here it was it was to be proven that AB by DE Is equal to BC by EF and all that so one we have already Sorry This one we have achieved Okay, so one ratio of this has been achieved Okay, AB by DE Is equal to AC by DF or this can be also written as CA by FD Right, this is what the third one we have achieved see here This too has been achieved Okay, this has been achieved now We had taken only one configuration. What we can do is we can repeat the process how we can now take a point here as let's say D dash And take another point here as E dash and join D dash E dash Similarly, you'll get the same result. You will be able to then prove that CA by FD is equal to CB by FE Right, so same the same process you repeat so we can always say similarly Yep, so you can you can say similarly Similarly, you can prove by taking appropriate Sides you can say similarly similarly Similarly You can say AC by DF AC by DF will be equal to BC by EF This you can prove And hence and hence all the three can be established together. So therefore We can say AB by BC AB by sorry AB by DE Is equal to BC by EF Is equal to CA by FD right So if the three angles are correspondingly equal Then the corresponding side ratios are also equal Correct. This is what is the first similarity criteria Now, let's say you would say what if this is possible this was possible the other way around that is AB is greater than DE Yeah, possible no problem But then do you think the proof is going to be changed only thing is you'll just swap the two diagrams and go again with the same Proof that means you will say that okay, this is bigger now in this case and this is smaller So instead of the second triangle you will cut the sides on the first triangle and do the same exercise Is it so doesn't matter whether AB is greater than DE or less than DE proof will not change Now the important fact to be noticed over here is The ASP comes into play, right? So if you know some of Some of three angles three angles Of a triangle Is how much 180 degrees, right? Therefore if two out of Three angles are equal three angles are correspondingly mark my words correspondingly correspondingly equal In two triangles that is so we are taking two triangles Triangles then you need don't check the third side then The third angle sorry because third angle has to be same because the total sum is 180 degrees, right? So only two angles if you match out of three that's good enough So if two out of three angles are correspondingly equal in two triangles then The triangles are similar triangles are Similar in the above case in the previous case we had to check all the three, right? But you don't need to check all the three why because let's say if you have two triangles like these One and let's say two these are similar. Let's say so if x is equal to this x And this y is equal to this y Okay, this y then don't you think the z third one has to be same? Right, why because the third angle this angle is 180 degrees minus x plus y So is the case here 180 degrees minus x plus y so this leads to another criteria, which is called a a criteria. So what is this called? This thing is called simply a a similarity criteria This is the corollary to the above criteria correct a a similarity criteria So what did we understand guys if two triangles are there? And two angles of each are correspondingly equal x is equal to x y is equal to y then the two triangles are Similar that is their sides will also be proportional Right there the ratio of the two sides corresponding sides Of the two triangle what ratio this by this And this by this And this by this side if you do that you'll get the same ratio. That's what is the first similarity criteria