In figure below, if $ \mathrm{AB} \| \mathrm{DC} $ and $ \mathrm{AC} $ and $ \mathrm{PQ} $ intersect each other at the point $ \mathrm{O} $, prove that $ \mathrm{OA} \cdot \mathrm{CQ}=\mathrm{OC} \cdot \mathrm{AP} $.
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Given:

\( \mathrm{AB} \| \mathrm{DC} \) and \( \mathrm{AC} \) and \( \mathrm{PQ} \) intersect each other at the point \( \mathrm{O} \).

To do:

We have to prove that \( \mathrm{OA} \cdot \mathrm{CQ}=\mathrm{OC} \cdot \mathrm{AP} \).

Solution:

In $\triangle A O P$ and $\triangle C O Q$,

$\angle A O P=\angle C O Q$             (Vertically opposite angles)

$\angle A P O=\angle C Q O$          (Alternate angles)

Therefore, by AA similarity,

$\triangle A O P \sim \triangle C O Q$

This implies,

$\frac{O A}{O C}=\frac{A P}{C Q}$      (Corresponding sides are proportional)

$O A \cdot C Q=O C \cdot A P$

Hence proved.

Tutorialspoint

Updated on: 10-Oct-2022

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