Two chords AB and CD of a circle intersect at a point outside the circle. Prove that
$(a) △ PAC ~ △ PDB (b) PA . PB = PC . PD$

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Solution

Given: AB and CD are two chords
To Prove:
$(a) △ PAC ~ △ PDB (b) PA . PB = PC . PD$

Proof:
$∠ ABD + ∠ ACD = 180^{\circ}$ .....(1)
(Opposite angles of a cyclic quadrilateral are supplementary)
$∠ PCA + ∠ ACD = 180^{\circ}$ ....(2)
(Linear Pair Angles)
Using (1) and (2), we get
$∠ ABD = ∠ PCA$
$∠ A = ∠ A$ (Common)
By AA similarity-criterion $△ PAC ~ △ PDB$
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.
$∴ \frac{PA}{PD} = \frac{PC}{PB} \Rightarrow PA . PB = PC . PD$

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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

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