In the given figure, AB and CD are two chords of a circle, intersecting each other at a point E.

Join AD

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 AXC subtends ∠AOC at the centre and ∠ADC at D on the circle.

∴ ∠AOC = 2∠ADC

$\Rightarrow \angle A D C=\frac{1}{2}(\angle A O C)$        ...(1)

Also, arc DYB subtends ∠DOB at the centre and ∠DAB at A on the circle.

∴ ∠DOB = 2∠DAB

$\Rightarrow \angle D A B=\frac{1}{2}(\angle D O B)$        ...(2)

Now, in ∆ADE,
∠AEC = ∠ADC + ∠DAB      (Exterior angle)

$\Rightarrow \angle A E C=\frac{1}{2}(\angle A O C+\angle D O B)$      (from (1) and (2))

Hence, $\angle A E C=\frac{1}{2}$ (angle subtended by arc $C X A$ at the centre $+$ angle subtended by arc $D Y B$ at the centre).