Data: ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively.
To Prove: PQRS is a rhombus.
Construction :
Diagonals AC and BD are drawn.
Proof:
In ∆ABC, P and Q are the mid-points of AD and BC.
∴ PQ || AC (Mid-point theorem)
PQ = \(\frac{1}{2}\)AC ………….. (i)
Similarly, in ∆ADC, S and R are the mid-points of AD and CD.
∴ SR || AC
SR = \(\frac{1}{2}\)AC …………… (ii)
Similarly, in ∆ABD,
SP || BD
SP = \(\frac{1}{2}\)BD ……………….. (iii)
Similarly, in ∆BCD,
QR || BD
QR = \(\frac{1}{2}\)BD ……………… (iv)
From (i), (ii), (iii) and (iv),
PQ = QR = SR = PS and Opposite sides are parallel.
∴ PQRS is a rhombus.