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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.

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Solution

## $In△ABC,\phantom{\rule{0ex}{0ex}}$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.

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