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Question
Prove that the rhombus, inscribed in a circle, is a square.
Solution
Let ABCD be a rhombus, inscribed in a circle
Now, ∠BAD + ∠BCD
(Opposite angles of a parallelogram are equal)
And ∠BAD + ∠BCD =180°
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
∴ ∠BAD + ∠BCD = `(180^circ)/2` = 90°
The other two angles are 90° and all the sides are equal.
∴ ABCD is a square.
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