Question

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

(i) ΔABC ~ ΔAMP

(ii) \(\frac{CA}{PA}\) = \(\frac{BC}{MP}\)

Answer 1

We have,

∠B = ∠M = 90°

And, ∠BAC = ∠MAP

In ΔABC and ΔAMP

∠B = ∠M (each 90°)

∠BAC = ∠MAP (Given)

Then, ΔABC  ~ ΔAMP (By AA similarity) 

So, CA/PM = BC/MP(Corresponding parts of similar triangle are proportional)