In the fig 6.39, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:
(i) ΔABC ~ ΔAMP
(ii) CA/PA = BC/MP
(i) In ΔABC and ΔAMP, we have
∠A = ∠A (common angle)
∠ABC = ∠AMP = 90° (each 90°)
∴ ΔABC ~ ΔAMP (By AA similarity criterion)
(ii) As, ΔABC ~ ΔAMP (By AA similarity criterion)
If two triangles are similar then the corresponding sides are equal,
Hence, CA/PA = BC/MP
