 Hello and welcome to the session let's discuss the following question it says prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medials. So let's now move on to the solution. Now we are given two similar triangles and we have to prove that the ratio of the areas of those two triangles is equal to the ratio, square of the ratio of their corresponding medials. So let A, B, C and P, Q, R, B to similar triangles and A, D and P, S are their corresponding medials. Now since A, D is median of triangle A, B, C and P, S is median of triangle P, Q, R therefore A, D is half of B, C and Q, S is half of Q, R and from this we have B, C is equal to this B, D is half of A, C so this implies B, C is double of B, D similarly Q, R is double of Q, S. Now triangle A, B, C is similar to triangle P, Q, R therefore ratio of their corresponding sides will be same that is A, B upon P, Q is equal to B, C upon Q, R is equal to C, A upon R, P. Now B, C is equal to twice of B, D and Q, R is equal to twice of Q, S so considering the first two we have A, B upon P, Q is equal to twice of B, D upon twice of Q, S. So this implies A, B upon P, Q is equal to B, D upon Q, S. Let's name this as one. Now we consider the triangles A, B, D and triangle P, Q, S. Now in triangle A, B, D and P, Q, S angle B is equal to angle Q, this is because triangle A, B, C is similar to triangle P, Q, R and since the two triangles are similar therefore their corresponding angles will be equal. Let's name this as two. Now from one and two we can say that triangle A, B, D is similar to triangle P, Q, S. This is by SAS similarity which says that if one angle of a triangle is equal to the one angle of other triangle and the sides including these angles are proportional then the two triangles are similar. Here we have seen that angle B is equal to angle Q and sides including these angles are proportional that is A, B upon P, Q is equal to B, D upon Q, S. So by SAS similarity criteria the two triangles are similar. Now again since the two triangles are similar therefore corresponding sides are in the same ratio. So we have A, B upon P, Q is equal to B, D upon Q, S is equal to A, D upon P, S. Let's name this as three. Now we have to prove that the ratio of the areas of two similar triangles is equal to the ratio square of the ratio of their corresponding mediums. That is we have to prove that the area of triangle A, B, C upon area of triangle P, Q, R is equal to the square of ratio of their corresponding mediums. That is area of triangle A, B, C upon area of triangle P, Q, R is equal to A, D square upon P, S square. Now again as triangle A, B, C is similar to triangle P, Q, R. Therefore ratio of the square of the areas that is we have to take the ratio of the areas that is area of triangle A, B, C upon area of triangle P, Q, R is equal to the square of ratio of corresponding sides that is it is equal to A, B square upon P, Q square is equal to B, C square upon Q, R square is equal to A, C square upon R, P square right. But from three we have A, B upon P, Q is equal to A, D upon P, S. So we have A, B square upon P, Q square is equal to A, D square upon P, S square. Let us name this as four. Now from three and four we have area of triangle A, B, C upon area of triangle P, Q, R is equal to A, D square upon P, S square. Hence we have proved that the ratio of the areas of two similar triangles is equal to the square of ratio of their corresponding medians. That is what we had to prove as the result is proved. So this completes the question and the session. Bye for now. Take care. Have a good day.