Question

In fig., seg AC and seg BD intersect each other at point P. `(AP)/(PC) = (BP)/(PD)`, then prove that ΔABP ~ ΔCDP

Answer 1

In ΔABP and ΔCDP,

`(AP)/(PC) = (BP)/(PD)`   ...[Given]

∠APB ≅ ∠CPD   ...[Vertically opposite angles]

∴ ΔABP ~ ΔCDP   ...[SAS test of similarity]