Question

$P$ and $Q$ are point of trisection of the diagnoal $BD$ of a parallelogram $ABCD$. Prove that $CQ||AP$.

Answer 1

Diagonals of parallelogram bisect each other.

BD and AC bisect at 0

$\Rightarrow AO=OC$ & $DO=OB\to \left(1\right)$

P & Q trisects diagnoal BD.

$DQ=QP=PB$

$DO=BO$

$PB=DQ$

$BO-PB=DO-DQ$

$OP=OQ\to \left(2\right)$

In Quadrilateral APC & AC and PQ bisect each other fraom (1) & (2). at O. so it in a parallelogram.

Hence , $CQ//AP$