$P$ 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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Updated on: 10-Oct-2022

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