 Hello and welcome to another session on triangles in this session We are going to discuss another very important theorem related to triangles and that is called external bisector theorem guys Okay, external bisector theorem which says that the external bisector of an angle of a triangle divides the opposite sides externally in the ratio of it should be side actually divide the externally opposite side Externally in the ratio of the sides Containing the angle first of all I will try and explain What does this mean and you've already seen the validation in the previous part of the video so the Given condition is this what is given condition the conditions which are given is Conditions which are there is given I've not mentioned the point e here. So let me okay Yeah, so this point is e Let's say this point is e the condition which is given is that angle C ad is equal to angle D a e all of you now know what is external angle of our triangle You know that this entire thing is the external angle. There are six external angle possible This is one angle which we are talking about and in and this C a e which is external angle C ad. Let me call it x and a e ad let me also call this as x. So these are Equal angles. We have to prove that what do we need to prove? to prove we need to prove that Like in the internal angle theorem a b Or let me write in the correct order. So order of writing also becomes helpful So be a upon a c you can see how I'm writing be a and then see so be a upon a c is BD Upon DC. So if you notice, how did I write a across like that? Right and then this B is B C and C right the writing also becomes little, you know So when you write in particular order becomes convenient to remember theorem as well So be a so basically I'm going like that be a then coming like that a C and similarly here It goes till D and then comes back to C Right, so this is how you're pleased remember like that. Okay, so this is what is the theorem Now we have to prove it. So let me just Remove all these unwanted Structures. So here it is BD by DC. Okay. Hope it is clear now again You see there are some, you know ratios which we are talking about. Okay, so what do we do then? We know that if there is a ratio or some ratios are involved Including the sides of the triangle then somehow we can use Basic proportionality theorem or the converse of it So where we need a triangle and need a parallel line the parallel line divides the given triangle in a given, you know the divide the sides containing the parallel line in a particular ratio, we know that so this is What was VPT? Let me just quickly discuss The VPT so this was basic proportionality theorem. There's a line. So AD upon AC is AE upon Sorry, this was B and this is C. So this was given that D is parallel to BC Then one is to two right is X is to Y Right, this is what we had learned. Okay, so we are going to use that. So for that I need some parallel line kind of structure here. So what do I do? and BC and CD should be involved Right, so D. So you're seeing the common points are A and D and there's one point C here So don't you think if I draw a line parallel like that? So let me construct CF And CF is parallel to AD. Okay, so what is the construction? construction construction is construction is CF parallel to AD drawn Right, this is what we did Okay, now, how is it going to help? Let's see now the moment I say CF is parallel to AD now Let's go to prove what can we say? CF is parallel to AD in Triangle B AD right CF is parallel to AD therefore, I can say BC upon CD Right, or we can actually use the corollary instead of writing BC upon CD Which all do we need? We need BD, right? So we need BD first. So BD upon BC will be BD upon DC corollary of BPT we are saying. So this whole By this will be equal to this whole by this Is that right? So hence we can say BD upon DC is equal to BA BA upon FA BA upon FA, clear? No problem. This is by BPT. Now Let me just take away this you all know what BPT is. So let me now talk about Triangle now if you see now CF is parallel to AD therefore, can I not say that This angle would be X So angle FCA is equal to angle CAD and what are they? They are nothing but alternate Interior angles since CF is parallel to AD, right? similarly Can I not say that? What else? Yeah, so this angle is also going to be X Is it why because corresponding on angles therefore similarly we can say angle AFC Will be equal to angle EAD and why is that CF is parallel to AD and they are corresponding angles Corresponding angles. I'm writing in short form, right? So that means in triangle AFC in Triangle AFC what happens AF is equal to AC why? because opposite Sides to Equal angle If you look closely AFC AFC is a isosceles triangle, isn't it? And why are we doing all of this? Because see in this particular Ratio, I got BD by DC So BD by DC is the RHS of what is to be proved and on the LHS of here. We have BA by FA So here BA is already available. We just need to prove somehow that FA is AC So we have to just prove that this part FA is equal to this part AC and that means These two angles have to be same, right? The opposite angles which I have shown as X must be same which actually is true because CF was parallel to AD so hence if you see the reverse order we could prove that We could prove that Triangle AFC is AFC is isosceles why because FCA was equal to CAD and AFC was equal to EAD and all of them were equal to X Isn't it? So hence AFC happens to be isosceles triangle the moment AFC is isosceles I can say therefore AF is equal to AC right opposite sides to which anyways I have written earlier as well so Therefore this let it be 1 and this let it be 2 So from 1 and 2 you can see we can write BD by DC is equal to BA upon AC Right BA upon AC and that is what we need to prove so hence proved Right. So this is what our target was. We have to prove BA by AC is Equal to BD by DC which is proved here, right? Also remember this theorem, right? So if an external angle is bisected, right? Then the angle bisector divides the opposite side in the Divide the opposite side externally in the ratio of the sides containing the angle Okay, this is what is external angle bisector theorem