(N/A) Let $ABCD$ be a quadrilateral such that the mid-points of sides $AB$,$BC$,$CD$,and $DA$ are $P$,$Q$,$R$,and $S$ respectively. We need to prove that the line segments joining the mid-points of opposite sides,i.e.,$PR$ and $SQ$,bisect each other at $O$.
Join $PQ$,$QR$,$RS$,and $SP$. Also,join $AC$ and $BD$.
In $\Delta ABC$,$P$ and $Q$ are the mid-points of $AB$ and $BC$ respectively.
By the Mid-point Theorem,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC$.
In $\Delta ADC$,$S$ and $R$ are the mid-points of $AD$ and $CD$ respectively.
By the Mid-point Theorem,$SR \parallel AC$ and $SR = \frac{1}{2} AC$.
From the above,$PQ \parallel SR$ and $PQ = SR$.
Since one pair of opposite sides of quadrilateral $PQRS$ is equal and parallel,$PQRS$ is a parallelogram.
We know that the diagonals of a parallelogram bisect each other.
Therefore,the diagonals $PR$ and $SQ$ of the parallelogram $PQRS$ bisect each other at $O$.
Hence,the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.