Question

11. Prove that the centre of the circle through A, B, C, D is the point intersection of its diagonals.​​

Answer 1

Answer:

Consider ABCD as a rectangle

We know that O is the point of intersection of the diagonals AC and BD

The diagonals of a rectangle are equal and bisect each other

So we get

OA=OB=OC=OD

We get O as the centre of the circle through A,B,C and D

Therefore, it is proved that the centre of the circle through A,B,C,D is the point of intersection of its diagonals.

Step-by-step explanation:

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Answer 2

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.

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