Question

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Answer 1

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.

Answer 2

Answer:

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