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Question

# The midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD are joined to form a quadrilateral. If AC = BD and AC ⊥ BD then prove that the quadrilateral formed is a square.

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Solution

## Given: In quadrilateral ABCD, AC = BD and AC ⊥ BD. P, Q, R and S are the mid-points of AB, BC, CD and AD, respectively. To prove: PQRS is a square. Construction: Join AC and BD. Proof: In ΔABC, $\because$ P and Q are mid-points of AB and BC, respectively. $\therefore$ PQ || AC and PQ = $\frac{1}{2}$AC (Mid-point theorem) ...(1) Similarly, in ΔACD, $\because$ R and S are mid-points of sides CD and AD, respectively. $\therefore$ SR || AC and SR = $\frac{1}{2}$AC (Mid-point theorem) ...(2) From (1) and (2), we get PQ || SR and PQ = SR But this a pair of opposite sides of the quadrilateral PQRS. So, PQRS is parallelogram. Now, in ΔBCD, $\because$ Q and R are mid-points of sides BC and CD, respectively. $\therefore$ QR || BD and QR = $\frac{1}{2}$BD (Mid-point theorem) ...(3) From (2) and (3), we get RS || AC and QR || BD But, AC ⊥ BD (Given) ∴ RS ⊥ QR But this a pair of adjacent sides of the parallelogram PQRS. So, PQRS is a rectangle. Again, AC = BD (Given) $⇒$ $\frac{1}{2}$AC = $\frac{1}{2}$BD $⇒$ RS = QR [From (2) and (3)] But this a pair of adjacent sides of the rectangle PQRS. Hence, PQRS is a square.

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