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Question
In the given figure, seg AC and seg BD intersect each other in point P and `"AP"/"CP" = "BP"/"DP"`. Prove that, ∆ABP ~ ∆CDP.
Solution
Given: Seg AC and seg BD intersect each other in point P and `"AP"/"CP" = "BP"/"DP"`.
To prove: ∆ABP ~ ∆CDP
Proof: In ∆ABP and ∆CDP,
`"AP"/"CP" = "BP"/"DP"` ...(Given)
∠APB ≅ ∠CPD ...(vertically opposite angles)
By SAS test of similarity,
∆ABP ~ ∆CDP
Hence Proved.
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