Question

Two triangles BAC and BDC, right angled at A and D respectively, are drawn on the same base BC and on the same side of BC. If AC and DB intersect at P, Prove that AP × PC = DP × PB .

Answer 1

Two Triangles are said to be similar if their i)corresponding angles are equal and ii)corresponding sides are proportional.(the ratio between the lengths of corresponding sides are equal)

•Similarity of triangles should be expressed symbolically using correct correspondence of their vertices
AA similarity: if two angles of one triangle are respectively equal to two angles of another triangle then the two Triangles are similar.

SOLUTION:

In ∆APB & ∆DPC,
∠A =  ∠D = 90°
∠APB = ∠DPC

[ vertically opposite angles]
∆APB  ~ ∆DPC
[By AA Similarity Criterion ]

AP/DP = PB/ PC
[corresponding sides of similar triangles are proportional]

AP × PC = DP × PB

Hence, proved.

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Answer 2

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