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.
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