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Show that the line segments joining the mid points of the opposite sides of a quadrilateral bisect each other.
To do:
We have to show that the line segments joining the midpoints of the opposite sides of a quadrilateral bisect each other.
Solution:
Let in a quadrilateral $ABCD$, $P, Q, R$ and $S$ are the midpoints of sides $AB, BC, CD$ and $DA$ respectively.
$PR$ and $QS$ intersect each other at $O$.
Join $PQ, QR, RS$ and $SP$ and join $AC$.
In $\triangle ABC$,
$P$ and $Q$ are mid-points of $AB$ and $BC$
This implies,
$PQ \parallel AC$ and $PQ = \frac{1}{2}AC$....…(i)
Similarly,
In $\triangle ADC$,
$S$ and $R$ are the mid-points of $AD$ and $CD$
This implies,
$SR \parallel AC$ and $SR = \frac{1}{2}AC$......(ii)
From (i) and (ii), we get,
$PQ = SQ$ and $PQ \parallel SR$
Opposite sides are equal area parallel.
This implies,
$PQRS$ is a parallelogram.
The diagonals of a parallelogram bisect each other.
Therefore,
$PR$ and $QS$ bisect each other.
Hence proved.
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