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Through the mid-point Q of side CD of a parallelogram ABCD, the line AR is drawn which intersects BD at P and produced BC at R. Prove that
(i) $AQ = QR$
(ii) $AP = 2PQ$
(iii) $PR = 2AP$
(i) $AQ = QR$
(ii) $AP = 2PQ$
(iii) $PR = 2AP$
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Sol. (i) $QC \parallel AB \therefore \triangle RQC \sim \triangle RAB$ (I Mark)
$\Rightarrow \frac{QR}{AR} = \frac{QC}{AB} = \frac{1}{2}$ (II Mark)
$\Rightarrow 2QR = AR \Rightarrow Q$ is the mid point of $AR$
$\therefore AQ = QR$
(ii) $\triangle PQD \sim \triangle PAB$ (III Mark)
$\therefore \frac{QP}{AP} = \frac{DQ}{BA} = \frac{1}{2}$ (IV Mark)
$\Rightarrow AP = 2PQ$
(iii) Since $AQ = QR$
$AP + PQ = PR - PQ$ (V Mark)
$\Rightarrow AP + \frac{1}{2}AP = PR - \frac{1}{2}AP$ (VI Mark)
$\Rightarrow PR = 2AP$
$\Rightarrow \frac{QR}{AR} = \frac{QC}{AB} = \frac{1}{2}$ (II Mark)
$\Rightarrow 2QR = AR \Rightarrow Q$ is the mid point of $AR$
$\therefore AQ = QR$
(ii) $\triangle PQD \sim \triangle PAB$ (III Mark)
$\therefore \frac{QP}{AP} = \frac{DQ}{BA} = \frac{1}{2}$ (IV Mark)
$\Rightarrow AP = 2PQ$
(iii) Since $AQ = QR$
$AP + PQ = PR - PQ$ (V Mark)
$\Rightarrow AP + \frac{1}{2}AP = PR - \frac{1}{2}AP$ (VI Mark)
$\Rightarrow PR = 2AP$