Question

In the given figure $ABCD$ is a parallelogram. $AC$ and $BD$ are the diagonals intersect at $O.P$ and $Q$ are the points of trisection of the diagonals $BD$. Prove that $CQ\parallel AP$ and also $AC$ bisects $PQ$

Answer 1

Consider the diagram given in the question.

As we know, the diagonals of a parallelogram bisect each other. Therefore,

$\Rightarrow AC$ and $BD$ bisect each other at point $O$.

Thus,

$\Rightarrow OA=OC$ and $OB=OD$

Now, consider points $P$ and $Q$.

$\Rightarrow BP=PQ=DQ$ ...... (1)

Now, since $OB=OD$, so

$PB+OP=OQ+DQ$

From equation (1),

$DQ+OP=OQ+DQ$

$\therefore OP=OQ$

Thus, $AC$ and $PQ$ bisects each other.

Thus, $APCQ$ is a parallelogram, because both the diagonals bisect each other. Also, since, $CQ$ and $AP$ are the opposite sides of the parallelogram $APCQ$, they are parallel to each other.

Hence proved.