Question

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.

Answer 1

Given: ABCD is a parallelogram and P is the midpoint of DC.

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

To prove: R is the midpoint of BC.

Construction: Join B and D and suppose it cut AC at O.

Proof: Now $OC=\frac{1}{2}AC$ (Diagonals of a parallelogram bisect each other) .......(1)

and CQ = $\frac{1}{4}AC$​​​​​​​ ...........(2)

From (1) and (2) we get

$CQ=\frac{1}{2}OC$​​​​​​​​​​​​​​

In ΔDCO, P and Q are midpoints of DC and OC respectively.

∴ PQ || DO (midpoint theorem)

Also in ΔCOB, Q is the midpoint of OC and PQ || AB

∴ R is the midpoint of BC (Converse of midpoint theorem)