Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Let ABCD is a quadrilateral and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively.

\(\Rightarrow \) AS = SD, AP = BP, BQ = CQ and CR = DR.

We have to show that: PR and SQ bisect each other i.e., SO = OQ and PO = OR.


Now, in \(\triangle{ADC}\), S and R are mid-point of AD and CD.

We know that, the line segment joining the mid-points of two sides of a triangle is parallel to the third side.

Thus, by midpoint theorem,
SR || AC and SR = (\(\frac{1}{2} \)) AC ...(i)

Similarly, in \(\triangle{ABC}\), P and Q are mid-point of AB and BC.


Thus, by midpoint theorem,
PQ || AC and PQ = (\(\frac{1}{2} \) ) AC ...(ii)
From Eq. (i) and (ii), we get,
PQ || SR and PQ = SR = (\(\frac{1}{2} \) ) AC

Therefore, Quadrilateral PQRS is a parallelogram whose diagonals SQ are PR.

Also, we know that diagonals of a parallelogram bisect each other. So, and bisect each other.
Hence, it is proved that PR and SQ bisect each other.