Question

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

Answer 1

In the given figure, ABCD is a quadrilateral such that A is the centre of the circle passing through B, C and D.

Join AC and BD.

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.

Here, arc CD subtends ∠CAD at the centre and ∠CBD at B on the circle.

∴ ∠CAD = 2∠CBD ...(1)

Also, arc CB subtends ∠CAB at the centre and ∠CDB at D on the circle.

∴ ∠CAB = 2∠CDB ...(2)

Adding (1) and (2), we get
∠CAD + ∠CAB = 2(∠CBD + ∠CDB)
⇒ ∠BAD = 2(∠CBD + ∠CDB)
⇒ ∠CBD + ∠CDB = $\frac{1}{2}$∠BAD

Hence, ∠CBD + ∠CDB = $\frac{1}{2}$∠BAD.