 Hello and welcome to the session. The given question says proof that in a right triangle the square of the hypotenuse is equal to the sum of the squares of the other two sites. Making use of the above, proof the following. We have to prove that in this figure, where a b c d is a rhombus, four times of a b square is equal to a c square plus b d square. So first let us prove the theorem. Here we have to prove that a c square is equal to a b square plus b c square where angle b is of 90 degree. So here we are given a right triangle a b c right angle dot b. And we have to prove that a c square is equal to a b square plus b c square. So first let us do a simple construction which we have already done here. We have to draw b d perpendicular on a d. Let us now start with the proof. Now in triangle a b c and triangle a d b we have angle a equal to angle a. This is common to both the triangles. And angle a b c is equal to angle a d b each 90 degrees. So this implies that triangle a b c is similar to triangle a d b by a similarity and this implies that a d by a b is equal to a b by a c. Since for two similar triangles sides are proportional. This further implies that a b square is equal to a d into a c. Similarly here a triangle c d b is similar to triangle c b a again by angle angle similarity. And this says that c d divided by b c is equal to b c divided by c a again. Since for two similar triangles their sides are proportional. This implies that b c square is equal to c d into c a. Let us denote this by equation number one and this by two. Now on adding equation number one and two on the left hand side we have a b square plus b c square. On the right hand side we have a d into a c plus c d into a c. Now on taking a c common we have inside the bracket a d plus c d. And this is equal to a c into a d plus c d is this distance which is a c. So we have a c square. Therefore from here we have that a c square is equal to a b square plus b c square hence proved. Now in this figure we have to show that four times of a b square is equal to a c square plus b d square. And this we should prove by using the Pythagoras theorem. Now we know that in a rhombus diagonal bisectat right angles right. Therefore here o c is equal to o a and d o is equal to b o. And also these four angles are of 90 degrees each. Since here we are given that a b c d is a rhombus. Now in triangle a o b which is right angle at o we have a b square is equal to a o square plus o b square. Which is the Pythagoras theorem which we have proved above right. And here a c can be written as a o plus o c and since the diagonal bisectature therefore a o plus o c can be written as a o. So here we have two times of a o which further implies that a c square is equal to four times of a o square. Similarly b d is equal to d o plus o b and again since diagonals bisect each other so we have d o is equal to o b. So here we can write two times of o b square. Sorry two times of o b. And on the square in both the sides we have b d square is equal to four times of o b square. Now on adding let us number this as one and this as two. We have left hand side as a c square plus b d square and on the right hand side we have four times of a o square plus four times of o b square. Which implies four times of a o square plus o b square and a o square plus o b square is a b square. Which is the Pythagoras theorem so we have four times of a b square. Hence we have four times of a b square equal to a c square plus b d square. So this completes the session. Bye and take care.