ABCD is parallelogram in which P is the midpoint of DC and Q is a point of AC such that CQ=1/4 AC. if PQ produced meets BC at R, prove that R is the midpoint of BC.

Given: ABCD is a parallelogram. P is the midpoint of CD.

Q is a point on AC such that CQ=(1/14)AC

PQ produced meet BC in R.

To prove: R is the midpoint of BC

Construction: join BD in O.Let BD intersect AC in O.

Prove: O is the midpoint of AC. {diagnols of a parallelogram bisect each other}

Therefore OC = (1/2) AC

⇒ OQ = OC-CQ = (1/2)AC - (1/4)AC = (1/4)AC.

⇒ OQ = CQ

therefore Q is the midpoint of OC.

In triangle OCD,

P is the midpoint of CD and Q is the midpoint of OC,

therefore PQ is parallel to OD (Midpoint theorem)

⇒ PR is parallel to BD

In triangle BCD,

P is the midpoint of CD and PR is parallel to BD,

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