Question

ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the midpoints of AB, AC, CD and BD respectively, show that PQRS is a rhombus.

Answer 1

$In△ABC,$P and Q are midpoints of AB and AC, respectively.
According to the midpoint theorem,
$PQ\parallel BC$ and PQ = $\frac{1}{2}$BC = $\frac{1}{2}$DA (It is given that AD = BC)

In $△$CDA, R and Q are midpoints of DC and AC, respectively
According to the midpoint theorem,
RQ$\parallel$DA and RQ = $\frac{1}{2}$DA

In $△$BDA, P and S are midpoints of AB and BD, respectively.
According to the midpoint theorem,
SP$\parallel$DA and SP = $\frac{1}{2}$DA

Similarly, in $△$CDB, R and S are midpoints of DC and BD, respectively.
According to the midpoint theorem,
SR$\parallel$BC and SR = $\frac{1}{2}$BC = $\frac{1}{2}$DA

Therefore, SP$\parallel$RQ, PQ$\parallel$SR and PQ = RQ = SP = SR

Hence, PQRS is a rhombus with all sides equal and opposite sides parallel to each other.