Question

Show that the figure formed by joining the midpoints of sides of a rhombus successively is a rectangle.

Answer 1

Let $ABCD$ be a rhombus and $P,Q,R$ and $S$ be the mid-points of sides $AB,BC,CD$ and $DA$ respectively.

In $△ABD$ and $△BDC$ we have

$SP\parallel BD$ and $SP=\frac{1}{2}BD$ ---- ( 1 ) [ By mid-point theorem ]

$RQ\parallel BD$ and $RQ=\frac{1}{2}BD$ ---- ( 2 ) [ By mid-point theorem ]

From ( 1 ) and ( 2 ) we get,

$SP\parallel RQ$

$PQRS$ is a parallelogram

As diagonals of a rhombus bisect each other at right angles.

$\therefore$ $AC\perp BD$

Since, $SP\parallel BD,PQ\parallel AC$ and $AC\perp BD$

$\therefore$ $SP\perp PQ$

$\therefore$ $\angle QPS={90}^o$

$\therefore$ $PQRS$ is a rectangle.