Question

$P$ and $Q$ are the points of trisection of the diagonal $BD$ of a parallelogram $ABCD.$ Prove that $CQ$ is parallel to $AP$. Prove also that $AC$ bisects $PQ$.

Answer 1

Since diagonals of a parallelogram bisect each other. Therefore,

$OA=OC$ and $OB=OD.$

Since $P$ and $Q$ are points of trisection of $BD.$

$\therefore BP=PQ=QD.$

Now, $OB=OD$ and $BP=QD$

$\Rightarrow OP-BP=OD-QD$

$\Rightarrow OP=OQ$

Thus, in quadrilateral $APCQ,$ we have

$OA=OC$ and $OP=OQ$

$\Rightarrow$ Diagonals of quadrilateral $APCQ$ bisect each other

$\Rightarrow APCQ$ is a parallelogram.

Hence, $AP\parallel CQ.$