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P and Q are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. Show that PQ is bisected at O.

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Solution.

Given: ABCD is a parallelogram whose diagonals bisect each other at O.
To Prove: PQ is bisected at O.


Proof: In \triangle ODP and \triangle OBQ
\angle BOQ=\angle POD {Vertically opposite angles}
\angle OBQ=\angle ODP {interior angles}
OB = OD {given}
 \triangle ODP\cong \triangle OBQ {by ASA congruence}
OP = OQ {by CPCT rule}
So, PQ is bisected at O

 

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