If ∆ABC and ∆AMP are two right triangles, right angled at B and M respectively such that ∠MAP = ∠BAC. Prove that

(1) It is given that $\triangle A B C$ and $\triangle A M P$ are two right angle triangles.

Now, in $\triangle A B C$ and $\triangle A M P$, we have

$\angle M A P=\angle B A C \quad$ (Given)

$\angle A M P=\angle B=90^{\circ}$

$\triangle A B C \sim \triangle A M P$ (AA Similarity)

(2) $\triangle A B C \sim \triangle A M P$

So, $\frac{C A}{P A}=\frac{B C}{M P}$  (Corresponding sides are proportional)