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

in given ∆ABP. & ∆CDP

AP/CP = BP/DP

angleBPA=angle DPC. (opposite angle)

from S.A.S similarity
∆ABP ~ ∆CDP

proved...

Answer 2

Answer:

Step-by-step explanation:

Given: Segment AC and BD  intersect each other at point P and  $\frac{AP}{CP}=\frac{BP}{DP}$

To Prove: ∆ABP is similar to ∆CDP.

Proof: From ∆ABP and  ∆CDP, we have

∠APB=∠DPC (Vertically opposite angles)

$\frac{AP}{CP}=\frac{BP}{DP}$ (Given)

Therefore, By SAS rule of congruency,

∆ABP is similar to ∆CDP

Hence proved.