ABCD is a quadrilateral in which AD = BC. If P, Q, R , S be the mid points of AB,

In △ABC, P and Q are mid points of AB and AC respectively.

So, $\mathrm{PQ} \| \mathrm{BC}$, and $\mathrm{PQ}=\frac{1}{2} \mathrm{BC}$           ..........(1)

Similarly, In $\triangle \mathrm{ADC}, \mathrm{QR}=\frac{1}{2} \mathrm{AD}=\frac{1}{2} \mathrm{BC}$    ............(2)

Now, In $\triangle \mathrm{BCD}, \mathrm{SR}=\frac{1}{2} \mathrm{BC}$         ........(3)

Similarly, In $\triangle \mathrm{ABD}, \mathrm{PS}=\frac{1}{2} \mathrm{AD}=\frac{1}{2} \mathrm{BC}$    .............(4)

Using (1), (2), (3) and (4)
PQ = QR = SR = PS
Since, all sides are equal
Hence, PQRS is a rhombus.