Question

In the adjoining figure, seg AC and seg BD intersect each other in point P and AP/CP = BP/DP Prove that, ∆ABP ~ ∆CDP 

Answer 1

Proof:

In ∆ABP and ∆CDP, 

AP/CP = BP/DP [Given] 

∠APB ≅ ∠CPD [Vertically opposite angles] 

∴ ∆ABP ~ ∆CDP [SAS test of similarity]