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Given that,
A cyclic quadrilateral .
ABCD,AC and BD$ are diameters of the circle where they meet at center O of the circle.
To prove:ABCD is a rectangle.
Proof: In triangle ΔAOD and ΔBOC,
OA=OC (both are radii of same circle)
∠AOD=∠BOC (vert.opp∠S)
OD=OB(both are radii of same circle)
∴ ΔAOD≅ΔBOC⇒AD=BC(C.P.C.T)
Similarly,by taking ΔAOB and ΔCOD,AB=DC
Also, ∠BAD=∠ABC=∠BCD=∠ADC=90°
(angle in a semicircle)
∴ ABCD is a rectangle.
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