In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) ∆ ABC ~ ∆ AMP
(ii) $\frac{CA}{PA} = \frac{BC}{MP}$

(i) In ΔABC and ΔAMP

∠ABC = ∠AMP = 90º

∠BAC = ∠MAP (Common angle)

Thus, ΔABC ∼ ΔAMP (AA criteria)

(ii) In ΔABC and ΔAMP

 $\frac{CA}{PA} = \frac{BC}{MP}$[∵ ΔABC ∼ ΔAMP]