Question

Prove that quadrilateral formed by joining the midpoints of consecutive sides of a rectangle is a rhombus.

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that ABCD be a rectangle such that P, Q, R, S are the midpoints of the sides of a rectangle AB, BC, CD, DA respectively.

Construction: Join PQ, QR, RS, SP and AC

Now, In triangle ABC

$\sf P \: is \: midpoint \: of \: AB$

$\sf Q \: is \: midpoint \: of \: BC$

So, By Midpoint Theorem, we get

$\implies\sf \: PQ \: \parallel \: AC \: and \: PQ \: = \: \dfrac{1}{2}AC - - - (1)$

Now, In triangle ADC

$\sf S \: is \: midpoint \: of \: AD$

$\sf R \: is \: midpoint \: of \: DC$

So, By Midpoint Theorem, we have

$\implies\sf \: SR \: \parallel \: AC \: and \: SR \: = \: \dfrac{1}{2}AC - - - (2)$

From equation (1) and (2), we concluded that

$\implies\sf \: PQRS \: is \: a \: parallelogram.$

[ ∵ If in a quadrilateral, one pair of opposite sides are equal and parallel, then quadrilateral is a parallelogram ]

$\implies\sf \: PQ = SR \: \: and \: \: PS = QR$

Now, As it is given that ABCD is a rectangle.

$\sf \: AD = BC$

$\sf \: \dfrac{1}{2} AD = \dfrac{1}{2} BC$

$\implies\sf \: AS = BQ$

Further also, P is the midpoint of AB

$\implies\sf \: AP = PB$

Now,

$\sf \: In \: \triangle \: ASP \: and \: \triangle \: QBP$

$\boxed{\begin{aligned}& \sf \:AS = BQ \: \{proved \: above \} \\ \\& \sf \: AP = PB \: \:\{proved \: above \} \\ \\& \sf \: \angle \: SAP= \angle \:PBQ \: \: \{each\: {90}^{ \circ} \} \end{aligned}} \implies\sf \: \triangle \: ASP \: \cong \: \triangle \: QBP$

[ By SAS Congruency ]

$\implies\sf \: SP = PQ$

$\implies\sf \: PQ = QR = RS = SP$

$\implies\sf \: PQRS \: is \: a \: rhombus$

Hence, the line segments joining the midpoints of the consecutive sides of a rectangle form a rhombus.