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In Fig, if PQRS is a parallelogram and $AB \parallel PS$ then prove that $OC \parallel SR.$

Answers (1)

To prove:-

$OC \parallel SR$

$\text{Proof:- In}$ $\Delta OPS \; and \; \Delta OAB$

$\angle POS=\angle AOB$ $\text{(common angle)}$

$\angle OSP=\angle OBA$ $\text{(corresponding angles)}$

$\therefore \Delta OPS \sim \Delta OAB$ $\text{(by AA similarity criterion)}$

$\Rightarrow \frac{OS}{OB}=\frac{PS}{AB}\; \; \; \; \; \; \; \; ....(i)$

$\Rightarrow \frac{QR}{AB}=\frac{CR}{CB}$

$\frac{PS}{AB}=\frac{CR}{CB}\; \; \; \; \; \; \; \; \; \; ....(2)$ $\left [ \because \; \text {PQRS is parallelogram,}\therefore PS=QR \right ]$

from equation (1) and (2)

$\frac{OS}{OB}=\frac{CR}{CB}\; or\; \frac{OB}{OS}=\frac{CB}{CR}$

Subtracting 1 from both sides

$\frac{OB}{OS}-1=\frac{CB}{CR}-1$

$\frac{OB-OS}{OS}=\frac{CB-CR}{CR}$

$\frac{BS}{OS}=\frac{BR}{CR}$

We know that if a line divides any two sides of a triangle in the same ratio then by converse of the basic proportionality theorem the line is parallel to the third side.

$\therefore SR\parallel OC$

Hence proved.

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