ntABCD is a parallelogram and P is the mid point of BC. Prove that AC and DP meet in a common point of trisect.n

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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. X is the mid-point of the side BC of a parallelogram ABCD. DX meets AC in Y Prove that

a r (Δ A D Y) = 2 a r (D Y C)

In the adjoining figure ABCD is a parallelogram. ‘P’ is a point on BC, DP and AB are produced meet at L. Prove that DP : PL = DC : BL.

$P$ is a point on side $B C$ of a parallelogram $A B C D$ . If $D P$ produced meets $A B$ produced at point $L$ , prove that :
(i) $D P : P L = D C : B L$ .
(ii) $D L : D P = A L : D C$ .

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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