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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)`
Theorem
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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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