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In figure below, if $ \mathrm{PQRS} $ is a parallelogram and $ \mathrm{AB} \| \mathrm{PS} $, then prove that $ \mathrm{OC} \| \mathrm{SR} $.
"
Given:
\( \mathrm{PQRS} \) is a parallelogram and \( \mathrm{AB} \| \mathrm{PS} \)
To do:
We have to prove that \( \mathrm{OC} \| \mathrm{SR} \).
Solution:
$Q R\|P S\| A B$ and $OC \| SR$
In $\triangle OPS$ and $\triangle OAB$,
$\angle POS = \angle AOB$ (Common)
$\angle OSP = \angle OBA$ (Corresponding angles)
Therefore, by AA similarity,
$\triangle OPS \sim \triangle OAB$
Using basic proportionality theorem, we get,
$\frac{PS}{AB} = \frac{OS}{OB}$.....…(i)
In $\triangle C Q R$ and $\triangle C A B$
$\angle Q C R=\angle A C B$
$\angle C R Q=\angle C B A$
Therefore, by AA similarity,
$\Delta C Q R \sim \triangle C A B$
This implies,
$\frac{Q R}{A B}=\frac{C R}{C B}$
$\frac{P S}{A B}=\frac{C R}{C B}$..........(ii) (Since $PS=QR$)
From (i) and (ii), we get,
$\frac{O S}{O B}=\frac{C R}{C B}$
$\frac{O B}{O S}=\frac{C B}{C R}$
On subtracting 1 from both sides, we get,
$\frac{O B}{O S}-1=\frac{C B}{C R}-1$
$\frac{O B-O S}{O S}=\frac{C B-C R}{C R}$
$\frac{B S}{O S}=\frac{B R}{C R}$
Therefore, by converse of basic proportionality theorem,
$S R \| O C$
Hence proved.