In figure seg AC and seg BD intersect each other in point P and AP/CP=

1.	In figure seg AC and seg BD intersect each other in point P and AP/CP=BP/DP.Prove that , ∆ABP ~ ∆CDP
Answer» ANSWER: In GIVEN ∆ACP & ∆BDP AP/CP = BP/DP Angle BPA =angle DPC. (OPPOSITE angle) from S.A.S similarity ∆ABP ~ ∆CDP proved...

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

Answer»

ANSWER:

In GIVEN ∆ACP & ∆BDP

AP/CP = BP/DP

Angle BPA =angle DPC. (OPPOSITE angle)

from S.A.S similarity

∆ABP ~ ∆CDP

proved...

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

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