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

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Solution

(i) In ΔPAC and ΔPDB,

$∠ P = ∠ P$ (Common)

$∠ P A C = ∠ P D B$ (Exterior angle of a cyclic quadrilateral is $∠ P C A = ∠ P B D$ equal to the opposite interior angle)

$∴ Δ P A C \sim Δ P D B$

(ii)We know that the corresponding sides of similar triangles are proportional.

$∴ \frac{P A}{P D} = \frac{A C}{D B} = \frac{P C}{P B}$

$\Rightarrow \frac{P A}{P D} = \frac{P C}{P B}$

$∴ P A . P B = P C . P D$

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Similar questions

In a circle, two chords AB and CD intersect at a point P inside the circle. Prove that

(a) $Δ P A C \sim Δ P D B$

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

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

Δ PAC~ Δ PDB

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

△ P A C \sim △ P D B

P A . P B = P C . P D

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