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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
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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