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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
(i) $∆PAC \sim ∆PDB$.
(ii) $PA \times PB = PC \times PD$.
"
Given:
Two chords $AB$ and $CD$ of a circle intersect each other at the point $P$ (when produced) outside the circle.
To do:
We have to prove that
(i) $∆PAC \sim ∆PDB$.
(ii) $PA \times PB = PC \times PD$.
Solution:
(i) In $\triangle \mathrm{PAC}$ and $\triangle \mathrm{PBD}$,
$\angle \mathrm{APC}=\angle \mathrm{BPD}$ (common angle)
$\angle \mathrm{ACP}=\angle \mathrm{ABD}$ (In a cyclic quadrilateral, an interior angle is equal to opposite exterior angle)
Therefore, by AA similarity,
$\triangle \mathrm{PAC} \sim \triangle \mathrm{PDB}$
Hence proved.
(ii) In $\triangle \mathrm{PAC}$ and $\triangle \mathrm{PBD}$,
$\angle \mathrm{APC}=\angle \mathrm{BPD}$ (common angle)
$\angle \mathrm{ACP}=\angle \mathrm{ABD}$ (In a cyclic quadrilateral, an interior angle is equal to opposite exterior angle)
Therefore, by AA similarity,
$\triangle \mathrm{PAC} \sim \triangle \mathrm{PDB}$
 This implies,
$\frac{\mathrm{PC}}{\mathrm{PB}}=\frac{\mathrm{PA}}{\mathrm{PD}}$
$\mathrm{PC} \times \mathrm{PD}=\mathrm{PA} \times \mathrm{PB}$
Hence proved.
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