Question

P,Q,R and S are respectively the mid-points of the sides AB ,BC, CD and DA of a quarilateral ABCD in which AC= BD, prove that PQRS is a rhombus.

Answer 1

if $AC=BD$

then $ABCD$ must be a square or a rectangle

if $ABCD$ is a rectangle,

$PB=\frac{x}{2},QB=\frac{y}{2}$

$\therefore PQ=\sqrt{P{B}^2+Q{B}^2}$ (Pythagorus theorem)

$=\sqrt{\frac{x^2}{4}+\frac{y^2}{4}}$

$PQ=\frac{1}{2}\sqrt{x^2+y^2}$

Similarly,

$AP=\frac{x}{2},AS=\frac{y}{2}$

$PS=\sqrt{A{P}^2+A{S}^2}=\sqrt{\frac{x^2}{4}+\frac{y^2}{4}}=\frac{1}{2}\sqrt{x^2+y^2}$

Similarly, all other side are equal

$\therefore PQRS$ will be a rhombus.