Question

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

Answer 1

Answer:

In given ∆ACP & ∆BDP

AP/CP = BP/DP

Angle BPA =angle DPC. (opposite angle)

from S.A.S similarity

∆ABP ~ ∆CDP

proved...