 Hello and welcome to the session. Let us discuss the following question. Question says prove that the sum of squares of the diagonals of parallelogram is equal to the sum of squares of its sides. First of all let us discuss basic idea behind the question. If we are given a triangle ABC and AB is a median of triangle ABC. Here clearly we can see ABC is a triangle and AD is a median on BC. Then AB square plus AC square is equal to 2AD square plus half BC square. We have already learnt this part in the previous question. This is the key idea to solve the given question. Now let us start with the solution. Let us consider a parallelogram ABCD whose diagonals AC and BD intersect each other at O. So we can write given a parallelogram ABCD whose diagonals intersect each other at O. Now we have to prove that sum of the squares of diagonals is equal to the sum of the squares of all its sides. So we have to prove that AC square plus BD square is equal to AB square plus BC square plus AD square plus AD square. Now let us start the required proof. First of all let us consider triangle ADC. In triangle ADC DO is a median on AC. We know the diagonals of parallelogram bisect each other. So DO is a median on AC since AO is equal to OC. This implies DO is a median on AC. Now by using the key idea we get AD square plus CD square is equal to 2OD square plus half AC square. Now let us name this expression as 1. Now we will consider triangle ABC. In triangle ABC DO is a median. We know the diagonals of parallelogram bisect each other. So DO is a median on AC in this triangle. Now by using key idea we get AB square plus BC square is equal to 2BO square plus half AC square. Now let us name this expression as 2 adding 1 and 2. We get AD square plus CD square plus AB square plus BC square is equal to 2DO square plus half AC square plus 2BO square plus half AC square. Now this can be written as AB square plus BC square plus CD square plus AD square is equal to we know half AC square plus half AC square is equal to AC square. And from these two terms we will take two common and get 2 multiplied by DO square plus BO square plus AC square on right hand side. Now we know DO is equal to BO is equal to half PD. We know diagonals of parallelogram bisect each other. So we will substitute half BD for DO and BO. So we get AB square plus BC square plus CD square plus AD square is equal to 2 multiplied by square of BD upon 2 plus square of BD upon 2 plus AC square now simplifying further we get AB square plus BC square plus CD square plus AD square is equal to 2 multiplied by 2 BD upon 2 whole square plus AC square. We know square of BD upon 2 plus square of BD upon 2 is equal to 2 multiplied by square of BD upon 2. Now this can be further written as AB square plus BC square plus CD square plus AD square is equal to 2 multiplied by 2 multiplied by BD square upon 4 plus AC square. Now we will cancel common factor 4 from numerator and denominator and we get AB square plus BC square plus CD square plus AD square is equal to BD square plus AC square or we can write it as AC square plus BD square is equal to AB square plus BC square plus CD square plus AD square. So this is our required answer. We were required to prove this only. This completes the session. Hope you understood the session. Take care and have a nice day.