Question

In the following figure, ABC and AMP are two right triangles, right-angled at B and M, respectively. Prove that:

(i) ΔABC ~ ΔAMP
(ii) $\frac{CA}{PA}=\frac{BC}{MP}$

Answer 1

1. In ∆ABC and ∆AMP,
   $\angle$ABC = $\angle$AMP (Each 90°)
   $\angle$A = $\angle$A (Common)
   ∴ ∆ABC ∼ ∆AMP (By AA similarity criterion)
2. $\Rightarrow\frac{CA}{PA}=\frac{BC}{MP}$