Points $P$ and $Q$ have been taken on opposite sides $A B$ and $C D$ , respectively of a parallelogram $A B C D$ such that $A P = C Q$ . Show that $A C$ and $P Q$ disect each other.

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Solution

In $△ A P O$ and $△ O Q C$

$∠ O A P = ∠ O C Q$ ( taking $A C$ as transversal then angles are on opposites sides of transversal)
$A P = C Q$ (given )
$∠ O P A = ∠ O Q C$ ( taking $P Q$ as transversal then angles are on opposites sides of transversal )

By $A S A$ criterion of congruency , $△ A P O$ is congruent to $△ O Q C$ .

$P O = O Q, A O = O C$ ( Corresponding parts of a congruent triangle are equal )
Hence $A C$ and $P Q$ bisect each other.

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