According to the question,
We have,
A quadrilateral ABCD such that A is the centre of the circle passing through B, C and D.
Construction:
Join CA and BD.
We know that,
In a circle, angle subtended by an arc at the center is twice the angle subtended by it at any other point in the remaining part of the circle
So,
The arc DC subtends ∠DAC at the center and ∠CAB at point B in the remaining part of the circle,
We get,
∠DAC = 2∠CBD …(1)
Similarly,
The arc BC subtends ∠CAB at the center and ∠CDB at point D in the remaining part of the circle,
We get,
∠CAB = 2∠CDB …(2)
From equations (1) and (2),
We have:
DAC + ∠CAB = 2∠CDB + 2∠CBD
⇒ ∠BAD = 2(∠CDB + ∠CBD)
⇒ 2(∠CDB + ∠CBD) = ½ (∠BAD)
