The diagonals of a quadrilateral ABCD are perpendicular. Show that the quadrilateral formed by joining the mid-points of its adjacent sides is a rectangle.

From figure,

ABCD is a quadrilateral in which diagonals AC and BD are perpendicular to each other. P, Q, R and S are mid-points of AB, BC, CD and DA.

In △ABC,

P and Q are mid-points of AB and BC respectively,

PQ || AC and PQ = $\dfrac{1}{2}$AC …….(1) (By mid-point theorem)

In △ADC,

S and R are mid-points of AD and DC respectively,

SR || AC and SR = $\dfrac{1}{2}$AC ………(2) (By mid-point theorem)

Using eqn. 1 and 2 we get,

PQ || SR and PQ = SR.

So, PQRS is a parallelogram.

In △ABD,

S and P are mid-points of AD and AB respectively,

SP || BD and SP = $\dfrac{1}{2}$BD ………(3) (By mid-point theorem)

Given,

AC and BD intersect at right angles,

From 3 we get,

SP || BD.

∴ SP ⊥ AC

From 2 we get,

SR || AC

∴ SP ⊥ SR i.e. ∠RSP = 90°.

∴ PQRS is a rectangle.

Hence, proved that the quadrilateral formed by joining the mid-points of its adjacent sides is a rectangle.