Let ABCD be a rhombus and P, Q, R and S be the mid-points of sides AB, BC, CD and DA, respectively (Fig.). Join AC and BD.
From triangle ABD, we have
SP = 1/2 BD and
SP || BD (Because S and P are mid-points)
Similarly, RQ = 1/2 BD and RQ || BD
Therefore, SP = RQ and SP || RQ
So, PQRS is a parallelogram. (1)
Also,AC ⊥ BD (Diagonals of a rhombus are perpendicular)
Further PQ || AC (From ∆BAC)
As SP || BD, PQ || AC and AC ⊥ BD,
therefore, we have SP ⊥ PQ, i.e. ∠SPQ = 90º. (2)
Therefore, PQRS is a rectangle[From (1) and (2)]
