Question

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

Answer 1

Join BD. Suppose it meets AC at S.
Since the diagonals of ||gm bisect each other, `CS=(1)/(2) AC`.
Now,` CS=(1)/(2) AC and CQ=(1)/(4) AC rArr CQ=(1)/(2) CS`.
`:. Q` is the midpoint of CS.
So, `PQ||DS` and therefore `QR||SB`.
In `Delta CSB, Q `is the midpoint of Cs and `QR||SB`, so R is the midpoint of BC.