A **secant** is a line that intersects a circle at two points. In the figure below, $latex \angle E$ is formed by two secants. The angle intercepts two arcs $latex \overline{AB}$ and $latex \overline{CD}$. In this post, we will prove that the measure of the angle formed by two secants intersecting outside a circle is half the difference of the arcs intercepted by it.

To prove this theorem we will connect $latex \overline{BC}$ and use the Inscribed Angle Theorem and Exterior Angle Theorem.

**Angle Secant Theorem**

The angle measure formed by two secants intersecting outside a circle is half the difference of the arcs intercepted by it

**Proof**

Draw $latex \overline{BC}$ and let the measures of arcs $latex AB$ and $latex CD$ be $latex \alpha$ and $latex \beta$ respectively. By the Inscribed Angle Theorem, the measure of the an angle inscribed in a circle is half the measure of its intercepted arc. Therefore, $latex \angle C = \frac{\alpha}{2}$ and $latex \angle CBD = \frac{\beta}{2}$.

Now, by the Exterior Angle Theorem,

$latex m \angle E + m \angle C = m \angle CBD$.

Substituting, we have t$latex m \angle E + \displaystyle \frac{\alpha}{2} = \displaystyle \frac{\beta}{2}$.

Therefore, $latex m \angle E = \displaystyle \frac{\beta}{2} – \displaystyle \frac{\alpha}{2}$

which is what we want to show.