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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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