Question

Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rectangle is a rhombus.

Answer 1

Let ABCD be the rectangle and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.
Join AC, a diagonal of the rectangle.
In ∆ ABC, we have:
∴ PQ ∣∣ AC and PQ = $\frac{1}{2}$AC [By midpoint theorem]
Again, in ∆ DAC, the points S and R are the mid points of AD and DC, respectively.
∴ SR ∣∣ AC and SR = $\frac{1}{2}$AC [By midpoint theorem]
Now, PQ ∣∣ AC and SR ∣∣ AC
⇒ PQ ∣∣ SR

Also, PQ = SR [Each equal to $\frac{1}{2}$ AC ] ...(i)
So, PQRS is a parallelogram.
Now, in ∆SAP and ∆QBP, we have:
AS = BQ
∠A = ∠B = 90^o
AP = BP
i.e., ∆SAP ≅ ∆QBP​
∴ PS = PQ ...(ii)
Similarly, ∆SDR ≅ ∆QCR​
∴ SR = RQ ...(iii)
From (i), (ii) and (iii), we have:
PQ = PQ = SR = RQ
Hence, PQRS is a rhombus.