Question

In the figure, ABCD is a parallelogram in which P is the mid -point 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 a mid -point of BC.

Answer 1

Given : In || gm ABCD,

P is the mid-point of DC and Q is a point on AC such that $CQ=\frac{1}{4}$AC. PQ is produced meets BC at R.

To prove : R is mid point of Bc

Constrction : Join BD

Proof : $\because$ In || gm ABCD,

$\because$ Diagonal AC and BD bisect each other at O

$\therefore$ AO = OC $=\frac{1}{2}AC$ ....(i)

In $\Delta OCD,$

P and Q the mid-points of CD and CO

$\therefore$ PQ || OD and $PQ=\frac{1}{2}OD$

In $\Delta BCD,$

P is mid -point of DC and PQ || OD (Proved above)

Or PR || BD

$\therefore$ R is mid -point BC.