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 center.

Answer 1

In $△BDC,\angle ADC$ is the exterior angle

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

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

By inscribed angle theorem, (an angle ${}^{\prime }{\theta }^{\prime }$ inscribed in a circle is half of the central angle ${}^{\prime }2{\theta }^{\prime }$ that subtends the same arc on the circle.)

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

From $eq.\left(1\right)$ and $\left(2\right),$ 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 center.