 Hi and how are you all today? The question says prove that the ratio of areas of two similar triangle is equal to the ratio of squares of their corresponding sides. Use the above in the following. In a trapezium A, B, C, D, O is the midpoint of intersection of A, C, and B, D. A, B are parallel to C, D and A, B is twice C, D. If the area of triangle A, O, B is 84 centimeter square, find the area of triangle C, O, D. So, for this purpose, let us first prove the given theorem. So, let's start with the solution of the first part of this question. That is, here we need to prove, or let us write first of all what is given to us. Here we are given that triangle A, B, C is similar to triangle D, E, F. We need to prove that area of triangle A, B, C upon area of triangle D, E, F is equal to A, B square upon D, E square, or it is equal to B, C square upon E, F square, or it is equal to A, C square upon D, F square. For this, we have drawn A, L perpendicular on V, C and D, M perpendicular on E, F, right? Now, let us start with our proof, right? Now, we are given that since these two triangles that is A, B, C is similar to triangle D, E, F, that means angle A is equal to angle T, angle B is equal to angle E and angle C is equal to angle F, right? Because these are the corresponding parts of the similar triangle, right? Further, this also shows that the ratio of the corresponding parts are equal to each other. That is also by CPS. Now, in triangle A, L, B and triangle T, M, E, we have angle B equal to angle E. That is from above, we have proved it. Angle A, L, B is equal to angle D, M, E, each being equal to 90 degree. So, we can say that therefore triangle A, L, B is similar to triangle D, M, E by A, A similarity. So, by this we can say that therefore A, L upon D, M is equal to A, B upon D, E, right? Let this be the first equation. Sorry, this be the second equation and let us suppose this is our first equation. Now, from the first and the second equation, we get A, B is equal to D, E is equal to A, L upon D, M is equal to B, C upon E, F is equal to A, C upon D, F, right? Now, we have area of triangle A, B, C equal to half into its base into its altitude. Area of triangle D, E, F is equal to half into its base that is E, F into its altitude, right? Let this be the third equation, this be the fourth equation and this be the fifth equation. Now, on dividing the fourth and the fifth equation, we get area of triangle A, B, C divided by area of triangle D, E, F is equal to half into B, C into A, L, divided by half into E, F into D, M. Now, in place of M, we can write B, C upon E, F, isn't it? So, it is equal to B, C, square upon E, F, square. Let this be the sixth equation. Now, on rewriting equation one, we have A, B upon D, E equal to B, C upon E, F is equal to A, C upon D, F or we can write that the squares are also equal to each other. Let this be the seventh equation. So, from the sixth equation and the seventh equation, we get that area of triangle A, B, C upon area of triangle D, E, F is equal to, it was equal to B, C, square and since B, C, square are equal to D, square can be written that it is equal to A, B, square upon D, E, square equal to B, C, square upon E, F, square equal to A, C, square upon D, F, square, right? So, we have, hence we have proved the given theorem as this completes the first part of the given question. Now, let's proceed with the solution of the second part. Now, here we have this trapezium A, B, C, D in which A, B is parallel to C, D and O is the point of intersection of A, C and B, D, right? So, let us write down what is given to us. We are given a trapezium A, B, C, D in which O is the point of intersection of A, C and also we are given that A, B is parallel to C, D and A, B is equal to twice C, D. And we are given area of triangle A, O, B is equal to 84 centimeter square. We need to find area of triangle C, O, D, right? So, let's start with our proof. Now, in triangle A, O, B and triangle C, O, we have angle 2 equal to angle 1 because they are forming alternate. We are given that these two lines are parallel to each other. So, when two lines, two parallel lines are intersected by a transversal then the alternate interior angles are equal to each other. Now, also angle 4 is equal to angle 3 because they are forming vertically opposite angles, right? So, by A is similarity triangle A, O, B is similar to triangle C, O, D. Now, since these two triangles are similar to each other, so we can write that area of triangle C, O, D upon area of triangle A, O, B is equal to C, D square upon A, B square. This is from the theorem that we have proved above that the ratio of the areas of two similar triangles are equal to the ratio of squares of their corresponding. Now, let us substitute the given values. We have area of triangle C, O, D upon 84 centimeter square equal to C, D square. Now, in place of A, D we can write 2 C, D. So, it will be 2 C, D whole square. Now, lastly we need to calculate its value now. So, we have area of triangle C, O, D equal to 84 into C, D square upon 4 into C, D square. So, we have thus the area of triangle C, O, D which was required to be found out in this question as 21 centimeter square. In this session, hope you understood the whole concept well. Have a nice day.