In the above figure ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that CQ = $\frac{1}{4}$ AC. If PQ produced meets BC at R, prove that R is the midpoint of BC.

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Solution

Consider the diagonal $B D$ as shown in the figure above.
$A C$ and $B D$ intersect at $O$ .

$P$ is the midpoint of $C D$ .
Hence, $\frac{C P}{C D} = \frac{1}{2}$

We know that, diagonals of a parallelogram bisect each other.
Hence, $O$ is the midpoint of $A C$

So, $O C = \frac{1}{2} A C$

Also, given that, $Q C = \frac{1}{4} A C$

Hence, $\frac{C Q}{C O} = \frac{1}{2}$

So, $\frac{C P}{C D} = \frac{C Q}{C O} = \frac{1}{2}$

Consider $Δ O C D$ and $Δ C Q P$ ,
$∠ C = ∠ C$

$\frac{C P}{C D} = \frac{C Q}{C O}$

Hence, by $S A S$ criteria, $Δ O C D \sim Δ C Q P$

So, $∠ O D C = ∠ Q P C$

$\Rightarrow P R | | B D$ [Corresponding angles are equal]

Hence, in $Δ C O B$ , by basic proportionality theorem,
$\frac{C R}{C B} = \frac{C Q}{C O} = \frac{1}{2}$

Hence, $R$ is the midpoint of $B C [H e n c e p r o v e d]$ .

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In the above figure ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that CQ =14 AC. If PQ produced meets BC at R, prove that R is the midpoint of BC.

In the above figure ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that CQ = $\frac{1}{4}$ AC. If PQ produced meets BC at R, prove that R is the midpoint of BC.