Let the vertex of an angle $\angle ABC$ be located outside a circle and let the sides of the angle intersect equal chords $AD$ and $CE$ with the circle. Prove that $\angle ABC$ is equal to half the difference of the angles subtended by the chords $DE$ and $AC$ at the centre.

(N/A) Let the vertex of $\angle ABC$ be $B$. The sides $BA$ and $BC$ intersect the circle at points $A, D$ and $C, E$ respectively such that chord $AD = CE$. Let $O$ be the centre of the circle. Join $OA, OC, OD, OE, AC$ and $DE$.
In $\Delta BAE$,the exterior angle $\angle DAE = \angle ABC + \angle AEC$ ... $(1)$
Since 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:
$\angle DAE = \frac{1}{2} \angle DOE$ ... $(2)$
Similarly,$\angle AEC$ is the angle subtended by arc $AC$ at the circumference,so $\angle AEC = \frac{1}{2} \angle AOC$ ... $(3)$
Substituting $(2)$ and $(3)$ in $(1)$:
$\frac{1}{2} \angle DOE = \angle ABC + \frac{1}{2} \angle AOC$
$\angle ABC = \frac{1}{2} \angle DOE - \frac{1}{2} \angle AOC$
$\angle ABC = \frac{1}{2} [\angle DOE - \angle AOC]$
Thus,$\angle ABC$ is equal to half the difference of the angles subtended by the chords $DE$ and $AC$ at the centre.