ABCD is a rectangle. Prove that the centre of the circle through A, B, C, D is the point of intersection of its diagonals.
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Solution
ANSWER: Given: ABCD is a cyclic rectangle whose diagonals intersect at O. To prove: O is the centre of the circle. Proof: Here, ∠BCD = 90° [Since it is a rectangle] So, BD is the diameter of the circle (if the angle made by the chord at the circle is right angle, then the chord is the diameter). Also, diagonals of a rectangle bisect each other and are equal. ∴ OA = OB = OC = OD BD is the diameter.∴ BO and OD are the radius. Thus, O is the centre of the circle. Also, the centre of the circle is circumscribing the cyclic rectangle. Hence, O is the point of intersection of the diagonals of ABCD.