In the figure,$AB$ and $CD$ are two chords of a circle intersecting each other at point $E$. Prove that $\angle AEC = \frac{1}{2}$ (angle subtended by arc $CXA$ at the centre $+$ angle subtended by arc $DYB$ at the centre).

(N/A) Given: $AB$ and $CD$ are two chords of a circle with centre $O$,intersecting at point $E$.
To prove: $\angle AEC = \frac{1}{2} (\angle AOC + \angle BOD)$.
Construction: Join $AC, BC, BD$ and $AD$.
Proof:
$1$. We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
$2$. Arc $CXA$ subtends $\angle AOC$ at the centre and $\angle ABC$ at the remaining part of the circle. Therefore,$\angle AOC = 2 \angle ABC$ $....(1)$
$3$. Similarly,arc $DYB$ subtends $\angle BOD$ at the centre and $\angle BCD$ at the remaining part of the circle. Therefore,$\angle BOD = 2 \angle BCD$ $....(2)$
$4$. Adding $(1)$ and $(2)$,we get: $\angle AOC + \angle BOD = 2(\angle ABC + \angle BCD)$ $....(3)$
$5$. In $\Delta CEB$,$\angle AEC$ is an exterior angle. By the exterior angle property,the exterior angle of a triangle is equal to the sum of the two interior opposite angles.
$6$. Therefore,$\angle AEC = \angle ABC + \angle BCD$ $....(4)$
$7$. From $(3)$ and $(4)$,we get: $\angle AOC + \angle BOD = 2 \angle AEC$.
$8$. Thus,$\angle AEC = \frac{1}{2} (\angle AOC + \angle BOD)$.
Hence,$\angle AEC = \frac{1}{2}$ (angle subtended by arc $CXA$ at the centre $+$ angle subtended by arc $DYB$ at the centre).