Question

ABCD is a rhombus and P Q R S are the midpoints of the sides ab BC CD respectively show that quadrilateral pqrs is a rectangle

Answer 1

 

Let us join AC and BD. 

In ΔABC, 

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

∴ PQ || AC and PQ = AC (Mid-point theorem) ... (1) 

Similarly in ΔADC, 

SR || AC and SR = AC (Mid-point theorem) ... (2) 

Clearly, PQ || SR and PQ = SR 

Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to 

each other, it is a parallelogram. 

∴ PS || QR and PS = QR (Opposite sides of parallelogram)... (3) 

In ΔBCD, Q and R are the mid-points of side BC and CD respectively. 

∴ QR || BD and QR =BD (Mid-point theorem) ... (4) 

However, the diagonals of a rectangle are equal. 

∴ AC = BD …(5) 

By using equation (1), (2), (3), (4), and (5), we obtain 

PQ = QR = SR = PS 

Therefore, PQRS is a rhombus

Answer 2

Hope it helps.

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