Question

Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.

Answer 1

Given: Let ABCD be a rectangle where P, Q, R, S are the midpoint of AB, BC, CD, DA.

To Prove: PQRS is a rhombus

Construction: Draw two diagonal BD and AC as shown in figure. Where BD = AC

(Since diagonal of the rectangle are equal)

Proof:

From ΔABD and ΔBCD 

PS = `1/2` BD = QR and PS || BD || QR

2PS = 2QR = BD and PS || QR                     ...(1)

Similarly, 2PQ = 2SR = AC and PQ || SR     ...(2)

From (1) and (2) we get

PQ = QR = RS = PS

Therefore, PQRS is a rhombus.

Hence, proved.