Question

△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 1

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]

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