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$P$ and $Q$ are the points of trisection of the diagonal $BD$ of the parallelogram $ABCD$. Prove that $CQ$ is parallel to $AP$. Prove also that $AC$ bisects $PQ$.
Given:
$P$ and $Q$ are the points of trisection of the diagonal $BD$ of the parallelogram $ABCD$.
To do:
We have to prove that $CQ$ is parallel to $AP$ and $AC$ bisects $PQ$.
Solution:
We know that,
Diagonals of a parallelogram bisect each other.
This implies,
$AO = OC$
$BO = OD$
$P$ and $Q$ are point of trisection of $BD$
Therefore,
$BP = PQ = QD$......…(i)
$BO = OD$.....…(ii)
Subtracting (ii) from (i), we get,
$BO - BP = OD - QD
$OP = OQ$
In quadrilateral $APCQ$,
$OA = OC$
$OP = OQ$
Diagonals $AC$ and $PQ$ bisect each other at $O$
Therefore,
$APCQ$ is a parallelogram
Hence, $AP \parallel CQ$.
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