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In the given figure, two chords $AB$ and $CD$ intersect each other at the point $P$. Prove that $AP \times PB = CP \times DP$.
"
Given:
Two chords $AB$ and $CD$ intersect each other at the point $P$.
To do:
We have to prove that $AP \times PB = CP \times DP$.
Solution:
In $\triangle \mathrm{APC}$ and $\triangle \mathrm{BPD}$,
$\angle \mathrm{APC}=\angle \mathrm{DPB}$ (Vertically opposite angles)
$\angle \mathrm{PAC}=\angle \mathrm{PDB}$ (Angles on the same segment are equal)
Therefore, by AA similarity,
$\triangle \mathrm{APC} \sim \triangle \mathrm{DPB}$
This implies,
$\frac{\mathrm{AP}}{\mathrm{PD}}=\frac{\mathrm{CP}}{\mathrm{PB}}$
$\mathrm{AP} \times \mathrm{PB}=\mathrm{CD} \times \mathrm{PD}$
Hence proved.
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