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Question
In the above figure, seg AC and seg BD intersect each other in point P. If `("AP")/("CP") = ("BP")/("DP")`, then complete the following activity to prove ΔABP ∼ ΔCDP.
Activity: In ΔABP and ΔCDP
`("AP")/("CP") = ("BP")/("DP")` ......`square`
∴ ∠APB ≅ `square` ......Vertically opposite angles
∴ `square` ∼ ΔCDP ....... `square` test of similarity.
Solution
In ΔABP and ΔCDP,
`("AP")/("CP") = ("BP")/("DP")` ......Given
∴ ∠APB ≅ ∠CPD ......Vertically opposite angles
∴ ΔABP ∼ ΔCDP ....... SAS test of similarity.
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