# In figure below, if $\mathrm{PQRS}$ is a parallelogram and $\mathrm{AB} \| \mathrm{PS}$, then prove that $\mathrm{OC} \| \mathrm{SR}$."

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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.

Updated on 10-Oct-2022 13:28:11