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Question
In the given figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that
(i) ΔPAC ∼ ΔPDB
(ii) PA.PB = PC.PD
Solution
(i) In ΔPAC and ΔPDB,
∠P = ∠P (Common)
∠PAC = ∠PDB (Exterior angle of a cyclic quadrilateral is ∠PCA = ∠PBD equal to the opposite interior angle)
∴ ΔPAC ∼ ΔPDB
(ii)We know that the corresponding sides of similar triangles are proportional.
∴ PA/PD = AC/DB = PC/PB
=>PA/PD = PC/PB
∴ PA.PB = PC.PD
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