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.

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Answer 1

Prove the statement.

Given: Vertex $B$ of angle $ABC$ lie outside the circle, chord $AD=CE$.

The angle subtended by an arc at the center is double the angle subtended by it any part of the circle.

From the given figure, the chord $DE$ subtends $\angle DOE$ at the center and $\angle DAE$ at the point $A$ on the circle.

Hence, $\angle DAE=\frac{1}{2}\angle DOE$…………$\left(1\right)$

From the given figure, the chord $AC$ subtends $\angle AOC$ at the center and $\angle AEC$ at the point $E$ on the circle.

Hence, $\angle AEC=\frac{1}{2}\angle AOC$…………$\left(2\right)$

In $∆ABE,\angle DAE$ is an exterior angle

$$\begin{gathered} \angle DAE=\angle ABC+\angle AEC \\ \Rightarrow \frac{1}{2}\angle DOE=\angle ABC+\frac{1}{2}\angle AOC \\ \Rightarrow \angle ABC=\frac{1}{2}\left(\angle DOE-\angle AOC\right) \end{gathered}$$

Hence it is proved that $\angle ABC$ is equal to half the difference of the angles subtended by the chords $AC$ and$DE$ at the center.