 Hello and welcome to the session. Let us understand the following question today. D and E are points on the sides C A and C B respectively of a triangle A B C, right angle that C. Prove that A E square plus B D square is equal to A B square plus D E square. Let us first see the diagram for this question. Here we have triangle A B C, right angle at C. We have taken points D and E on A C and B C respectively. And here we have joined these points D E, B D and A E. Now let us write the solution. In a right angle triangle A C E by Pythagoras theorem, we can write A E square that is our hypotenuse is equal to A C square plus C E square. Let us name this equation as 1. Also in right angle triangle D C B by Pythagoras theorem, D B square that is the hypotenuse is equal to some of the squares of other two sides that is D C square plus D C square. Let us name it as 2. Now adding 1 and 2, it implies A E square plus B D square is equal to A C square plus C E square. A E square plus B C square plus B C square. Now it implies A E square plus B D square is equal to A C square plus B C square C square plus B C square. Now by Pythagoras theorem in triangle A B C, we have A B square is equal to A C square plus B C square and in triangle D C E, we have D E square is equal to B C square plus C E square. So this equation becomes A E square plus B D square is equal to A C square plus B C square can be replaced by A B square plus C E square plus D C square can be replaced by D E square. And this is our required result. Hence proved, if you have understood the question, bye and have a nice day.