Question

If $ABCD$ is a parallelogram and $AP=CQ$, show that $AC$ and $PQ$ bisect each other.

Answer 1

Given: $ABCD$ is a parallelogram and $AP=CQ$.

Let $M$ be the point of intersection of $AC$ and $PQ$.

To show: $AC$ and $PQ$ bisect each other.
Proof:
In $\Delta AMP$ and $\Delta CMQ$.
$\angle MAP=\angle MCQ$
(Alternate interior angles with $AC$ as transversal)
$AP=CQ$ (Given)
$\angle APM=\angle CQM$
(Alternate interior angles with $PQ$ as transversal)
$\therefore \Delta AMP\cong \Delta CMQ$ (By ASA congruence rule)
$\Rightarrow AM=CM\cdots \left(i\right)$ (By C.P.C.T. rule)
and $PM=MQ\cdots \left(ii\right)$ (By C.P.C.T. rule)
From (i) and (ii) we can say that $M$ is the mid-point of $PQ$ and $AC$.
Therefore, $AC$ and $PQ$ bisect each other.
Hence proved.