So, we have, a quadrilateral ABCD where AD = BC

And P, Q, R and S are the mid-point of the sides AB, AC, and BD.

We need to prove that PQRS is a rhombus.

In ΔBAD, by midpoint theorem we get, 

PS||AD and PS = 1/2 AD…………(i)

InΔCAD, by midpoint theorem we get, 

QR||AD and QR = 1/2 AD …………..(ii)

Compare (i) and (ii)

PS||QR and PS = QR

Since one pair of opposite sides is equal and parallel,

Then, we can say that PQRS is a parallelogram…………(iii)

Now, In ΔABC, by midpoint theorem

PQ||BC and PQ = 1/2 BC…………..(iv)

And AD = BC …………………………..(v)

Compare equations (i) (iv) and (v), we get, 

PS = PQ ………………………………….(vi)

From (iii) and (vi), we get, 

PS = QR = PQ 

Since PQRS is a parallelogram, PQRS is a rhombus.