Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same sides of BC. If AC and DB intersect at P, prove that AP × PC = BP × PD.

Question

Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same sides of BC. If AC and DB intersect at P, prove that AP &#x00d7 PC  = BP &#x00d7 PD.

Philoid · Accepted Answer

Given: Two right triangles ABC and BDC on the same hypotenuse BC. AC and BD intersect at P.
To prove: AP × PC = BP × PD
Proof: In ΔABP and ΔDCP
∠A = ∠D (= 90°) (given)
∠APB = ∠DPC (vertically opposite angles)
∴ ΔABP ~ ΔDCP (AA similarity axiom)
∴ AB = BP = AP (corresponding sides of similar Δ s are proportional)………………….(1)
DC CP DP
From (1) BP = AP
CP DP
By cross multiplication,
BP × DP = AP × PC (proved).