In the adjoining figure, ABCD is a ||gm in which P is the midpoint of DC and Q is a point on AC such that `CQ=(1)/(4)AC.` Also, PQ when produced meets BC at R. Prove that R is the midpoint of BC.

Explanation

Join BD, intersecting AC at O.
Then, `AO=OC " "[ because " diagonals of a ||gm bisect each other"].`
`therefore CQ=(1)/(4) AC=(1)/(4)xx(2OC)=(1)/(2)OC.`
Thus, Q is the midpoint of OC.
Now, in `triangle CDO`, P and Q are the midpoint of CD and CO respectively.
`therefore ` PQ||DO and thererfore, QR||OB [`because " PQ||DO" rArr ` PQR||DOB].
Now, in `triangle COB`, Q is the midpoint of CO and QR||OB.
`therefore ` R must be the midpoint of BC.