Question

in fig. seg AC and seg BD intersect each other at point P and AP/CP =BP/DP.prove that, traingle ABP similar traingle 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

Step-by-step explanation:

in in triangle APB and triangle CDP

1. AP/CP = BP/DP

2.ANGLE BPA = ANGLE BPC REASON :OPPOSITE ANGLE

3. THEREFORE TRIANGLE APB ~ TRIANGLE CDP( BY SAS TEST)