If PQRS is a parallelogram and AB || PS, then prove that OC || SR.
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Given: In $\triangle\text{ABC},$ O is any point in the interior of $\triangle\text{ABC}.$ OA, OB, OC are joined. PQRS is a parallelogram such that P, Q, R and S lies on segments OA, AC, BC and OB and PS || AB.To prove: OC || SR
Proof: In $\triangle\text{OAB}$ and $\triangle\text{OPS}$
$\text{PS}\parallel\text{AB}$ [Given]
$\therefore\angle1=\angle2$ [Corresponding angles]
$\therefore\angle3=\angle4$ [Corresponding angles]
$\therefore\triangle\text{OPS}\sim\triangle\text{OAB}$ [by AA similarity criterion]
$\Rightarrow\frac{\text{OP}}{\text{OA}}=\frac{\text{OS}}{\text{OB}}=\frac{\text{PS}}{\text{AB}}\ .....(\text{i})$
PQRS is a parallelogram so PS || QR. .......(ii)
QR || AB ....... (iii) [from (i), (ii)]
In $\triangle\text{CQR}$ and $\triangle\text{CAB},$
$\text{QR }\parallel\text{ AB}\ .....(\text{iii}) $
$\therefore\angle\text{CAB}=\angle5$ [Corresponding angle]
$\therefore\angle\text{CBA}=\angle6$ [Corresponding angle]
$\therefore\triangle\text{CQR}\sim\triangle\text{CAB}$ [by AA similarity criterion]
$\Rightarrow\frac{\text{CQ}}{\text{CA}}=\frac{\text{CR}}{\text{CB}}=\frac{\text{QR}}{\text{AB}}$
PQRS is a parallelogram.
$\therefore\text{PS }\parallel\text{ QR}$
$\therefore\frac{\text{PS}}{\text{AB}}=\frac{\text{CR}}{\text{CB}}=\frac{\text{CQ}}{\text{CA}}\ ......(\text{iv})$
$\Rightarrow\frac{\text{CR}}{\text{CB}}=\frac{\text{OS}}{\text{OB}}$ [from (i) and (iv)]
These are the ratios of two sides of $\triangle\text{BOC}$ and are equal so by converse of BPT, SR || OC.
Hence proved.
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