Prove that the quadrilateral obtained by joining the mid-points of an isosceles trapezium is a rhombus.

P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.

Join AC and BD.

Since, ABCD is an isosceles trapezium

Its diagonals are equal.

AC = BD = x (let)

Now, in ∆ABC

P and Q are the mid-points of AB and BC

So, PQ || AC and PQ = $\dfrac{1}{2}$AC = $\dfrac{1}{2}$x (By midpoint theorem) … (i)

Similarly, in ∆ADC

S and R mid-point of AD and CD

So, SR || AC and SR = $\dfrac{1}{2}$AC = $\dfrac{1}{2}$x (By midpoint theorem) … (ii)

From (i) and (ii), we have

PQ || SR and PQ = SR = $\dfrac{1}{2}x$ …….(iii)

In ∆CBD,

R and Q are the mid-points of CD and BC

So, QR || BD and QR = $\dfrac{1}{2}$BD = $\dfrac{1}{2}$x (By midpoint theorem) … (iv)

Similarly, in ∆ABD

S and P mid-point of AD and AB

So, SP || BD and SP = $\dfrac{1}{2}$BD = $\dfrac{1}{2}$x (By midpoint theorem) … (v)

From (iv) and (v), we have

QR || SP and QR = SP = $\dfrac{1}{2}x$ …….(vi)

From (iii) and (vi) we get,

PQ = QR = SR = SP and QR || SP, PQ || SR.

So, sides of PQRS are equal and opposite sides are parallel.

Hence, proved that PQRS is a rhombus.