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Question

# ABCD is a kite where AB =AD and BC =CD .P,Q,R and S are the mid -points of the sides AB, BC, CD and AD respectively. Prove that PQRS is a rectangle.

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Solution

## Given: ABCD is a kite where AB =AD and BC =CD P,Q,R and S are the mid -points of the sides AB, BC, CD and AD respectively. To prove : PQRS is a rectangle. Construction: Join AC and BD, which intersect in O. Proof: In ΔABD,P and S are the mid-points of the AB and AD respectively. PS||BD and PS=12BD...(1)(Mid point theorem) In ΔBCD,Q and R are the mid-points of the sides BC and CD respectively. QR||BD and QR=12BD....(2) (Mid point theorem) From (1) and (2), PS||QR and PS =QR ∴ PQRS is a parallelogram ...(3) (In a quadrilateral, if one pair opposite sides is parallel and equal, then it is a parallelogram.) Since ABCD is a Kite, ∴∠AOD=90∘...(4) (In kite, diagonals intersect each other at 90∘) Since PS||BD FS||OE ....(5) SR||AC (Using mid -point theorem) ∴SE||OF Front (5) and (6), we get OESF is a parallelogram (Both pair of opposite sides are parallelogram) ∠FOE=∠FSE=90∘ (Opposite angles of parallelogram are equal). ∴∠PSR=90∘...(7) From (3) and (7), we can conclude that Parallelogram ABCD is a rectangle.

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