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)