 Hello friends, welcome to the session, I am Alka. Let's discuss triangles. Our given question is in figure 6.39 ABC and AMP are two right triangles, right angle at B and M respectively. Proof that triangle ABC is similar to triangle AMP and CA upon PA equal to BC upon MP. Here is the given figure with A, B, C and AMP as two right angles, right angle at M and B. Now let's start with the solution. We are given triangle ABC triangle AMP right triangles. We are also given that the triangle ABC and AMP are right angles at B and M. So we can write as angle M equal to angle B equal to 90 degree. Now we have to show triangle ABC is similar to triangle AMP. Now let's begin with the proof. In triangle ABC triangle AMP we have angle BC equal to angle AMP equal to 90 degree. We are given B, A, C, B, A, C equal to angle. Therefore we can say that by A, A criteria of similarity we have triangle ABC similar to triangle AMP. Hope you understood the first part. Now let's see the next part. We have to show that CA upon PA equal to BC upon MP. Now let's start with the solution. Now as we have already seen in the first part that triangle ABC is similar to triangle AMP proved above. Therefore now we all know that if two triangles are similar then the ratio of the corresponding sides are equal. So we can say that for two similar triangles the ratio equal. This implies that CA upon PA equal to BC upon MP is proved. Hope you understood the solution of both the parts and enjoyed them. See you and goodbye.