question_answer

In given figure, ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that\[CQ=\frac{1}{4}AC\] and 4 PQ produced meet BC at R, then

A) \[CR=\frac{1}{3}CB\]               

B)         CR= RB         

C)        \[CR=\frac{1}{2}RB\]               

D)         None of these

Correct Answer: B

Solution :

Join B and D. Suppose AC and BD intersect at O. Then. \[OC=\frac{1}{2}AC\] Now, \[CQ=\frac{1}{4}AC\][Given] \[\Rightarrow \]\[CQ=\frac{1}{2}OC\] In \[\Delta COD,P\]and Q are the mid-points of DC and OC respectively. \[\therefore \]\[PQ|DO\]               [By mid-point theorem] Also, in \[\Delta COB,Q\]is the mid-point of OC and \[QR||OB\] \[\therefore \] R is the mid-point of BC. [By converse of mid-point theorem] \[\Rightarrow \]CR = RB