Question

$ABCD$ is a rectangular and $P,Q,R$ and $S$ are mid points of the sides $AB,BC,CD$ and $DA$ respectively. Show that the quadrilateral $PQRS$ is a rhombus .

Answer 1

REF.Image

Given - P,Q,R,S are the midpoint of AB,AC,CD,AD respectively.

ABCD is a rectangle

to prove :- PQRS is rhombus.

proof :-

$BP=AP=\frac{1}{2}AB$

$DS=AS=\frac{1}{2}AD$

$DR=RC=\frac{1}{2}DC$

$CQ=BQ=\frac{1}{2}BC$

$\therefore AP=BP=RC=DR=x$

& $AS=DS=CQ=BQ=y$

In $\Delta APS$

$\therefore \angle A={90}^{\circ }$( given by property of root)

$\therefore SP=\sqrt{x^2+y^2}$ (by Pythagoras)

Similarly

$SR=\sqrt{x^2+y^2},RQ=\sqrt{x^2+y^2},PQ=\sqrt{x^2+y^2}$

$\therefore SP=PQ=RQ=RS$

Hence SPQR is a parallelogram - (1)

$PO=AS=y$

& $SO=AP=x$

In $\Delta POS$

$O{P}^2+O{S}^2=S{P}^2$

Hence $\angle SOP={90}^{\circ }$

Hence, diagonal of parallelogram PQRS is intersect at ${90}^{\circ }$ hence it is rhombus.