 Hello and welcome to the session. Let us understand the following question today. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. Now let us first see the figure. Here we have A, B, C, D as a rhombus with O as a center and diagonal intersecting at O. Now let us write the solution. Let the diagonals A, C and B, D of rhombus A, B, C, D intersect each other at O. Now since the diagonals of a rhombus bisect each other at right angles therefore angle A, O, B is equal to angle B, O, C is equal to angle C, O, D is equal to angle D, O, A is equal to 90 degree. And therefore A, O is equal to O, C and B, O is equal to O, D since it is also bisecting the diagonals. So now triangle A, O, B this triangle is a right triangle right angled at therefore by Pythagoras theorem AB square is equal to OA square plus OB square which implies AB square is equal to OA square is half of AC square plus OB square is half of DD square because DA is equal to OC sorry OA is equal to OC and OD is equal to OB. Now it implies AB square is equal to AC square by 4 plus BD square by 4. Now which implies taking 4 on this side we get 4AB square is equal to AC square plus BD square. Let us name this equation as 1. Similarly we can write 4 BC square is equal to AC square plus BD square let it be 2 and 4 CD square is equal to AC square plus BD square let it be 3 and 4 AD square is equal to AC square plus BD square let it be 4. So now adding 1, 2, 3 and 4 we get it implies 4AB square plus 4BC square plus 4CD square plus 4AD square is equal to AC square plus BD square plus AC square plus BD square plus AC square plus BD square plus AC square plus BD square which implies taking four commons so we get four multiplied by AB square plus BC square plus CD square plus AD square which is equal to four AC square plus four VD square. Now which implies four multiplied by AB square plus BC square plus CD square plus AD square is equal to four multiplied by AC square plus VD square. Now we see that this four gets cancelled with this four so we are left with AD square plus BC square plus CD square plus AD square is equal to AC square plus VD square and this is our desirable which says that sum of the squares of the sides of the rhombus is equal to the sum of the squares of the diagonals. This proved. I hope you understood the question. Bye and have a nice day.