 Welcome friends. Let's take up another theorem related to parallelograms. This is another property of a parallelogram and it says the diagonals of a parallelogram bisect each other. So what does it say? It says that diagonals of a parallelogram bisect each other. That means wherever they meet they split each other into two equal parts. So let's try to prove this. So let me take this parallelogram A, B, C, D and let's draw the diagonals because that's what we have to deal with. So this is diagonal B, D and this one is this one is A, C okay and let the point of intersection be called O. Okay now we have to prove that A O is equal to O C and D O is equal to O B. So let's do the proof. So what is given? So you can start with given. Given is A, B, C, D is a parallelogram. So you can write that and the moment you say so automatically other things are assumed. What are what are assumed? So A, B is parallel to C, D and A, C sorry not A, C, A, D is parallel to B, C. So these two things are there. Okay and you have to prove to prove. What is to be proven? A O is equal to O C and B O is equal to O D. That's what is meant by bisection right. So hence A, C is bisecting O B, D B and D B is bisecting O C. So you have to notify here. So this is A O is equal to O C and with two such lines, D O is equal to O B right. This is what we have to prove. How to go about it? So let's try the proof. So whenever such conditions are there where we have to prove equality of two sides, one way of doing it is trying to find congruences right. So if you are able to prove that the given parts are corresponding parts of congruent triangles then our job is done and that's where it all starts. So let's say let us try to figure out which two triangles should be taken so that all these components which are to be proved are there in them. So there are two choices. One is if you take A O D this one and this one. Let's say these two triangles. But in these two triangles, the problem is O C is not involved. So again it will be difficult to prove. Do we have a better choice? We do have. So let's say if we choose these two triangles then yes, all the components which are there to be proved are involved. So let's try. So we say in triangle A O D and triangle C O B. Mind you again, I'm not writing anything but C O B because that is where the correspondence is achieved. So angle A in the given triangle must be equal to angle C in likewise. You'll see why in a short while. Now D O D A O D A O this angle. Let's say X is equal to this angle is equal to XY alternate interior angles. D A O is equal to angle B C O and you can write alternate interior angles. Right? Secondly, the other angle is also equal and what is that? Let's say this angle is equal to this angle. Isn't it? Let it be Y. Y. So hence I say angle A D O is equal to angle C B O is equal to Y and this was equal to X and because of the same reason alternate interior angles. Correct? And we know that A D is equal to B C. Why? Opposite sides of a parallelogram. We just proved in previous session that opposite sides of a parallelogram are equal. Therefore, by which congruence criterion, you know by A S A congruence, congruence criterion, what do we prove? We get triangle A O D is congruent to triangle C O B. Again, the correspondence, the order of these vertices are very, very important because A must be equal to angle C. Right? Angle A in this angle A O D must be equal to angle C C O B. Right? So angle O must be equal to O and D must be equal to B like that in the given two triangles. So hence correspondence is very, very important. Okay. Now, yes. So hence what can we say? Now clearly A O will be equal to C O and you can write congruent parts of sorry corresponding parts of congruent triangle C P C T. Similarly, similarly, you can write D O is equal to O B. Again, C P C T hence proved. This is what we needed to prove. Isn't it? So hence now you know how to prove such theorem. So hence, the strategy was to find out one, two triangles which are, which include all those sides and then using the properties previously proved, you can prove they are congruent part of, sorry, corresponding parts of two congruent triangles and hence you can achieve this proof.