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  • ABCD is such a quadrilateral such that A is the centre of the circle passing through B, C and D. Prove that \angle CBD + \angle CDB =\frac{1}{2}\angle BAD.

ABCD is such a quadrilateral that A is the centre of the circle passing through B, C and D. Prove that  \angle CBD + \angle CDB =\frac{1}{2}\angle BAD.

Answers (1)

Given: A circle passing through points B, C, D and ABCD is a quadrilateral having a centre at A.

To prove: \angle CBD +\angle CDB =\frac{1}{2}\angle BAD

Construction: Join AC & BD

Proof:

We know that the angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle

Arc DC subtends \angleDAC at the centre and \angleDBC at point B in the remaining part of the circle.

\therefore \angleDAC = 2\angleCBD … (i)

Similarly, arc BC subtends ÐBAC at the centre and ÐBDC at point D in the remaining part of the circle.

\therefore \angleCAB = 2 \angleCDB … (ii)

On adding equations (i) and (ii)

We get

\angleDAC + \angleCAB = 2 \angleCBD + 2 \angleCDB

\angleDAB = 2 (\angleCBD + \angleCDB)

\angle CBD +\angle CDB =\frac{1}{2}\angle BAD

Hence proved

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