In the figure, ABCD is a parallelogram in which P is the mid -point of DC and Q is a point on Ac such that $C Q = \frac{1}{4} A C$ If PQ produced meets BC at R, prove that R is a mid -point of BC.

Q. ABCD is a parallelogram as shown in figure. If AB=2AD and P is mid-point of AB, then

∠ C P D

is equal to

Q. A parallelogram ABCD has P the mid- point of DC and Q intersects AC such that

C Q = \frac{1}{4} A C

. PQ produced meets BC at R prove that :
(a) R is the mid-point of BC
(b)

P R = \frac{1}{2} D B

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In a quadrilateral ABCD, the point P divides DC in the ratio 1:2 and Q is the mid point of AC. If AB+2AD+BC−2DC=kPQ, then k is equal to:

The correct option is A −6Given AB+2AD+BC−2DC=kPQ⇒AB+2AD+BC−2DC=AC+2AD−2DC=AC+2(AC+CD)−2DC=3AC−4DC=3(2QC)−4(32PC)=6QC−6PC=6(QC+CD)∴kPQ−6QP−6PQ ( Given )⇒k=−6.Hence, the answer is −6.

In a quadrilateral $A B C D,$ the point $P$ divides $D C$ in the ratio $1 : 2$ and $Q$ is the mid point of $A C$ . If $A B + 2 A D + B C - 2 D C = k P Q,$ then $k$ is equal to: