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Question
In Quadrilateral ABCD, side AD || BC, diagonal AC and BD intersect in point P, then prove that `"AP"/"PD" = "PC"/"BP"`
Solution
Proof: seg AD || seg BC and BD is their transversal. ...[Given]
∴ ∠DBC ≅ ∠BDA ...[Alternate angles]
∴ ∠PBC ≅ ∠PDA ...(i)[D−P−B]
In ΔPBC and ΔPDA,
∠PBC ≅ ∠PDA ...[From (i)]
∠BPC ≅ ∠DPA ...[Vertically opposite angles]
∴ ΔPBC ∼ ΔPDA ...[AA test of similarity]
∴ `"BP"/"PD" = "PC"/"AP"` ...[Corresponding sides of similar triangles]
∴ `"AP"/"PD" = "PC"/"BP"` ...[By alternendo]
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