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Question
In fig., seg AC and seg BD intersect each other at point P. `(AP)/(PC) = (BP)/(PD)`, then prove that ΔABP ~ ΔCDP
Theorem
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Solution
In ΔABP and ΔCDP,
`(AP)/(PC) = (BP)/(PD)` ...[Given]
∠APB ≅ ∠CPD ...[Vertically opposite angles]
∴ ΔABP ~ ΔCDP ...[SAS test of similarity]
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