Inscribed Angles
Prove and apply the Inscribed Angle Theorem: an inscribed angle is half its intercepted arc. Identify corollaries (angles inscribed in a semicircle are right; inscribed angles intercepting the same arc are congruent).
A lesson of Geometry, in Circles, about 10 minutes.
By the end of this lesson, you will be able to apply the Inscribed Angle Theorem, prove that inscribed angles intercepting the same arc are congruent, state and use Thales' theorem (angle in a semicircle = 90°), and find angle measures in inscribed polygons.
The Big Idea
An inscribed angle is always exactly half its intercepted arc. This single theorem has profound consequences: all inscribed angles that intercept the same arc are equal to each other (regardless of where on the circle the vertex sits), and any angle inscribed in a semicircle is a right angle. These are among the most elegant results in plane geometry.
Inscribed Angle Theorem
An inscribed angle is half the measure of its intercepted arc.
$$m\angle ABC = \frac{1}{2} m\overset{\frown}{AC}$$
where $B$ is on the circle and $\overline{AB}$, $\overline{BC}$ are chords.
Proof (for center inside the angle): Draw diameter through $B$ and center $O$. Two isosceles triangles form (all radii equal). Use the exterior angle theorem for each, sum them.
Corollaries
Corollary 1: Inscribed angles that intercept the same arc are congruent.
$$\angle BAC \cong \angle BDC \quad \text{(both inscribed, same arc } \overset{\frown}{BC}\text{)}$$
Corollary 2 (Thales' Theorem): An inscribed angle that intercepts a semicircle (diameter) is a right angle.
$$\overline{BC} \text{ is a diameter} \implies \angle BAC = 90°$$
Proof: Semicircle = 180° arc; inscribed angle = $\frac{1}{2}(180°) = 90°$.
Corollary 3: Opposite angles of an inscribed quadrilateral (cyclic quadrilateral) are supplementary.
$$\angle A + \angle C = 180° \quad \text{and} \quad \angle B + \angle D = 180°$$
Proof: $\angle A = \frac{1}{2}\overset{\frown}{BCD}$ and $\angle C = \frac{1}{2}\overset{\frown}{BAD}$. Since $\overset{\frown}{BCD} + \overset{\frown}{BAD} = 360°$, we get $\angle A + \angle C = 180°$.
Example 1 — Basic Inscribed Angle
$m\overset{\frown}{AB} = 140°$. Find the inscribed angle intercepting $\overset{\frown}{AB}$.
$$m\angle ACB = \frac{1}{2}(140°) = 70°$$
Example 2 — Same Arc, Multiple Angles
$\angle ADB = 35°$, with $D$ on circle $O$. Find $\angle ACB$ where $C$ is also on circle $O$.
Both angles intercept $\overset{\frown}{AB}$ → $\angle ACB = \angle ADB = 35°$.
Example 3 — Thales' Theorem
$\overline{AB}$ is a diameter. $\angle AEB = ?$
By Thales' theorem, $\angle AEB = 90°$ for any point $E$ on the circle.