△ABC and △AMP are two right triangles right angled at B and M resp

1.	△ABC and △AMP are two right triangles right angled at B and M respectively. Prove that (i) △ABC ~ △AMP (ii) \(\frac{CA}{PA}=\frac{BC}{MP}\)
Answer» Given: △ABC; ∠B = 90° AAMP; ∠M = 90° R.T.P : i) △ABC ~ △AMP Proof: In △ABC and △AMP ∠B = ∠M [each 90° given] ∠A = ∠A [common] Hence, ∠C = ∠P [∵ Angle sum property of triangles] ∴ △ABC ~ △AMP (by A.A.A. similarity) ii) △ABC ~ △AMP (already proved) \(\frac{AB}{AM}=\frac{BC}{MP}=\frac{CA}{PA}\) [∵ Ratio of corresponding sides of similar triangles are equal] \(\frac{CA}{PA}=\frac{BC}{MP}\)

△ABC and △AMP are two right triangles right angled at B and M respectively. Prove that (i) △ABC ~ △AMP (ii) \(\frac{CA}{PA}=\frac{BC}{MP}\)

Answer»

Given: △ABC; ∠B = 90°

AAMP; ∠M = 90°

R.T.P : i) △ABC ~ △AMP

Proof: In △ABC and △AMP

∠B = ∠M [each 90° given]

∠A = ∠A [common]

Hence, ∠C = ∠P

[∵ Angle sum property of triangles]

∴ △ABC ~ △AMP (by A.A.A. similarity)

ii) △ABC ~ △AMP (already proved)

\(\frac{AB}{AM}=\frac{BC}{MP}=\frac{CA}{PA}\)

[∵ Ratio of corresponding sides of similar triangles are equal]

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