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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» 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) |
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