Question

A rectangle abcd has mid points p,q,r,s of ab,bc,cd and da. prove that pqrs is rhombus

Answer 1

Let the rectangle have side lengths AB = 2x and BC = 2y.

Then PB = x and BQ = y. So, Pythagoras gives us that PQ = √(x^2 + y^2).

Similarly, QC = DS = SA = y and AP = CR = RD = x so by Pythagoras,

QR = RS = SP = √(x^2 + y^2).

Hence as PQ = QR = RS = SP, PQRS must be a rhombus, as required.