Question

In parallelogram ABCD, P is the mid-point of DC. Q is a point on AC such that CQ = `(1)/(4)"AC"`. PQ produced meets BC at R. Prove that

(i) R is the mid-point of BC, and

(ii) PR = `(1)/(2)"DB"`.

Answer 1

(i) Join B and D. Suppose AC and BD cut at O. Then,

OC = `(1)/(2)"AC"`

Now, 
CQ = `(1)/(4)"AC"`

⇒ CQ = `(1)/(2)"OC"`

In ΔDCO, P and Q are the mid-points of DC and OC respectively.
∴ PQ || DO
Also, in ΔCOB, Q is the mid-point of OC and PQ || OB
Therefore, R is the mid-point of BC, R being PQ produced.

(ii) In ΔBCD, P and R are the mid-points of DC and BC respectively.
Also PR || BD

Therefore, PR = `(1)/(2)"BD"`.