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Question
In the following figure, ABC and AMP are two right triangles, right-angled at B and M respectively, prove that:
- ΔABC ~ ΔAMP
- `("CA")/("PA") = ("BC")/("MP")`
Solution 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.]
Solution 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]
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