 Hello friends, so welcome again to another session on theorems of triangles and its angles Now we have taken one corollary of the previous angles and property which we saw in the previous session the corollary says if the bisectors of angles ABC and ACB of a triangle ABC need at a point B Okay, so you can see bisectors of angle ABC Which is BD and bisector of angle ACB which is CD meet at point D then angle BDC, okay, BDC is this angle you can see that is given by beta is equal to 90 degrees plus half angle B a AC which is given as alpha So the corollary is given here. So beta is 90 degree plus half angle alpha take 30 seconds pause and understand the figure After that what we'll do is we will try to validate this and in different configuration of the same triangle will try to validate Now indeed beta, which is angle BDC is equal to 90 degrees plus half angle alpha, which is BAC Okay, so first let's validate it and then in the later part, let's try to prove this So what I'm going to do is I'm going to change the configuration of triangle ABC and In all the cases in all situations, we must have this particular relation always true Okay, so I'm going to change the position of a so here I go So you can see I am changing and I'm now here at this point So again, you can see beta is 90 degrees plus half angle alpha in this case alpha is 60.41 degrees So half of 60.41 plus 90 is 120.21 degrees. Okay, so Wherever I take B the relation always holds. There is nowhere Where it doesn't hold is it I can change the location of B as well see I'm changing the location of B as well But then in that case also this angle is 90 degree plus twice this angle Isn't it? So if I change C as well The relation is going to be true. So at any configuration you can see you can pause the video and see that This particular relationship always holds now we will look at the proof of this Corollary what is a corollary corollary is nothing but Statement in mathematics drawn from a particular theorem. So that's a corollary. Okay, so next let's prove this particular So here is yet another theorem which says that if the bisectors of triangle ABC and ACB of triangle ABC meet at the point B then Angle BDC is equal to 90 degrees plus half angle BAC Okay, I simply read the statement. Let us now try and understand what this means We just saw the validation by the way in the previous part of this video where in GeoGibra We established this particular term that is indeed if you take an triangle ABC and This angle or BD. Let's say is the divisor or BD is the Angle bisector of angle B and CD is angle bisector of angle C. So if this angle is Alpha This angle is beta this angle also is alpha and This angle is also beta. That is what is the meaning of angle bisectors. Now we had just Saw in the previous sessions angle some properties. So we are going to use that property to prove this so angle BDC What is BDC? So let me call it gamma BDC is gamma and let's say this angle is delta So don't get confused if I'm using such Greek letters. So alpha beta gamma delta So let's see the proof now So how do we do this proof? first of all the The customary step given what is given BD and CD are Angle bisectors Angle bisectors of what? angle ABC and Sorry, I have to write angle ABC So this is angle ABC and Angle ACD Correct and the word respectively Respectively why because of this one-on-one mapping that is ABC's bisector is BD And ACD is bisector is CD Okay, so to prove To prove Angle BDC is equal to 90 degrees plus half Angle BAC Correct or BDC is or you have to prove what gamma is equal to 90 degrees Plus half Delta This is what we have to prove. Let's try and prove now We don't need a particular construction here. So you can say in So I'm saying in triangle in triangle ABC in triangle ABC, what can I say? Delta The angle A that is Plus twice angle alpha Plus twice angle beta. So three angles put together Delta twice alpha is you can see this is twice alpha Angle B and twice beta is angle C. No problem in that This is how much 180 degrees and what is the reason behind my writing such things? Angle some Property of a triangle correct Isn't it now this let it be equation number one Next is I'm writing here so that everything is in one frame second. I can write in a Attention to triangle BDC BDC so same angle some property will hold here as well. So gamma plus Alpha plus beta is equal to 180 degrees Isn't it? Yeah, what is this again ASP? I'm writing in short form ASP angle some property of a triangle and I can do some manipulation over here. What manipulation can I do? I can just multiply this by two So multiplying this whole equation by two Why did I do this? You will come to know Just in a while Two alpha plus two beta is equal to three sixty degrees Is it it? Let it be two Okay, now we will do one operation. What is that? Either or you can say two alpha plus two beta. Let's say what is it? 360 degrees minus two gamma Is it it? right and from one From one what can we say? Two alpha plus two beta is equal to 180 degrees minus delta right So this is let's say three This is let's say four So you can say from three and four the LHS of three and four Are same if you see correct. So we can equate the RHS as well So 360 degrees minus two gamma Is equal to 180 degrees minus delta Is it it? So If I simplify I will get Two gamma Is equal to 360 degrees minus 180 degrees plus delta Okay now this is Nothing but two gamma is equal to 180 degrees Plus delta right so what will be gamma guys? So i'm going to multiply this equation by half or divide by two whichever way you want to say it and this is equal to gamma Is equal to 90 degrees plus delta by two correct? And this is where gamma is equal to 90 degree plus delta by two and this is what we needed to prove check Right, this is what we achieved. Okay, so please remember this corollary as well