Question

Let the vertex of an 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 $AC$ and $DE$ at the centre.

Answer 1

$\Rightarrow$In $△BDC$,

$\Rightarrow$$\angle ADC$ is the exterior angle

$\therefore \angle ADC=\angle DBC+\angle DCB$ .... (1)

(The measure of an exterior angle is equal to the sum of the remote interior angles)

$\Rightarrow$By inscribed angle theorem,

$\Rightarrow$$\angle ADC=\frac{1}{2}\angle AOC$ and $\angle DCB=\frac{1}{2}\angle DOE$ ....... (2)

From (1) and (2), we have

$\Rightarrow$$\frac{1}{2}\angle AOC=\angle ABC+\frac{1}{2}\angle DOE$ ...[Since $\angle DBC=\angle ABC$]

$\Rightarrow \angle ABC=\frac{1}{2}(\angle AOC-\angle DOE)$

Hence, $\angle ABC$ is equal to half the difference of angles subtended by the chords $AC$ and $DE$ at the centre.