Question

The diagonals of a quadrilateral ABCD are perpendicular to each other. Prove that the quadrilateral obtained by joining the midpoints of its adjacent sides is a rectangle.

Answer 1

Given:

In quad. ABCD, diagonals \[AC \perp BD\]. Let P, Q, R, S be the midpoints of sides AB, BC, CD, DA, respectively.

\[\text{In } \triangle ABC \text{ and } \triangle ADC \text{,}\]

\[PQ \parallel AC \text{ and } PQ = \frac{1}{2} AC \text{ (By Mid-point Th.)}\]

\[SR \parallel AC \text{ and } SR = \frac{1}{2} AC \text{ (By Mid-point Th.)}\]

Therefore, PQ || SR and PQ = SR. Since one pair of opposite sides is equal and parallel, PQRS is a parallelogram.

\[\text{In } \triangle ABD \text{,}\]

\[PS \parallel BD \text{ (By Mid-point Th.)}\]

We have \[PQ \parallel AC\] and $PS \parallel BD$. Since $AC \perp BD$, and $PQ \parallel AC$ and $PS \parallel BD$, Therefore, $PQ \perp PS$. Thus, \[\angle QPS = 90^\circ\].

Since parallelogram PQRS has one right angle (\[\angle QPS = 90^\circ\]), it is a rectangle. (Hence proved.)