Question

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

Answer 1

In $△DPC$ and $△BQA$

$DP=BQ$

$AB=CD$

$\angle CDP=\angle ABQ$

$△DPC\cong △BQA$

$\therefore CP=AQ,AP=CQ$

$⟹APCQ$ is a parallelogram

$⟹AP\parallel CQ$