Question

A parallelogram ABCD has P the mid- point of DC and Q intersects AC such that $CQ=\frac{1}{4}AC$. PQ produced meets BC at R prove that :
(a) R is the mid-point of BC
(b) $PR=\frac{1}{2}DB$

Answer 1

Given ABCD is a parallelogram and P is midpoint of DC

also, $CQ=\frac{1}{4}AC$

To Prove : R is mid point of BC

Proof : Now

$OC=\frac{1}{2}AC$ (Diagonals of parallelogram bisect each other) ...(i)

and $CD=\frac{1}{4}AC$ ...(ii)

From (i) and (ii)

$CD=\frac{1}{2}OC$

In $\Delta DCO$ $P$ and $Q$ are midpoint of DC and OC Respectively

$\therefore PQ\parallel DO$

Also in $\Delta COB$ Q is midpoint of OC and $PQ\parallel DB$

$\therefore R$ is midpoint of $BC$

$\therefore$ in $\Delta ABCPR\parallel DB$

$\frac{CP}{CD}=\frac{CR}{CB}=\frac{PR}{BD}$

$\frac{PR}{DB}=\frac{1}{2}$

$\therefore PR=\frac{1}{2}DB$