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

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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.PB = PC.PD

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Similar questions

Δ PAC~ Δ PDB

△ P A C \sim △ P D B

P A . P B = P C . P D

(a) △ PAC ~ △ PDB (b) PA . PB = PC . PD

Two chords AB and CD of a circle intersect at a point P outside the circle. Prove that

(1) $Δ P A C \sim Δ P D B$

(2) $P A . P B = P C . P D$ .

(a) △ PAC ~ △ PDB (b) PA . PB = PC . PD

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