In the adjoining 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 the adjoining figure, seg AC and seg BD intersect each other in point P and AP/CP = BP/DP Prove that, ∆ABP ~ ∆CDP
In the adjoining figure, seg AC and seg BD intersect each other in point P and AP/CP = BP/DP Prove that, ∆ABP ~ ∆CDP
Proof:
In ∆ABP and ∆CDP,
AP/CP = BP/DP [Given]
∠APB ≅ ∠CPD [Vertically opposite angles]
∴ ∆ABP ~ ∆CDP [SAS test of similarity]