Question

19. In fig., ABCD is a parallelogram in which
P is the mid-point of DC and Q is a pointon AC such that CQ =1/4AC. If PQ produced meets BC at R, prove that R is a
mid-point of BC.​

Answer 1

Answer:

Given: ABCD is a parallelogram. P is the mid point 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 mid point of BC

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

Prove : O is the mid point of AC. {diagnols of 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 mid point of OC.

In triangle OCD,

P is the mid point of CD and Q is the mid point of OC,

therefore PQ is parallel to OD (Mid point theorem)

=> PR is parallel to BD

In traingle BCD,

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

therefore R is the mid point of BC (Converse of midmidpoint