Question

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

1. ΔABC ~ ΔAMP
2. `("CA")/("PA") = ("BC")/("MP")`

Answer 1

(i) In ΔABC and ΔAMP,

∠ABC = ∠AMP        ...(Each 90°)

∠A = ∠A                 ...(Common)

∴ ΔABC ∼ ΔAMP            ...(By AA similarity criterion)

(ii) 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]

∴ `"CA"/"PA"="BC"/"MP"`          ....[Corresponding parts of similar Δ are proportional.]

Answer 2

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]

∴ `"CA"/"PA"="BC"/"MP"`          ....[Corresponding parts of similar Δ are proportional]