An inscribed angle has its vertex on the circle and chords as its sides. It intercepts the arc that lies inside it. Compare it with the central angles of Central Angles and Arcs, whose vertex is at the center.
Why it works
Start with the case where one side of the inscribed angle $\angle APB$ is a diameter, $\overline{PB}$ through the center $O$. Then $OP = OA$ (radii), so $\triangle OPA$ is isosceles and its base angles are equal; call each $x$ (this is Isosceles and Equilateral Triangles). The central angle $\angle AOB$ is an exterior angle of $\triangle OPA$, so it equals $x + x = 2x$. The arc $\widehat{AB}$ measures $2x$ and the inscribed angle measures $x$: half. Any other inscribed angle can be split by the diameter through $P$ into two angles of this kind (or written as their difference), so the rule holds every time.
Key idea
Inscribed Angle Theorem: an inscribed angle measures half of its intercepted arc. So inscribed angles that intercept the same arc are congruent, and an angle inscribed in a semicircle is a right angle.
Worked example: From the arc
Inscribed $\angle QPR$ intercepts $\widehat{QR}$, which measures $112^\circ$. Find $m\angle QPR$.
- An inscribed angle is half its arc: $m\angle QPR = \frac{112}{2} = 56^\circ$.
Answer: $56^\circ$
Worked example: Same arc, same angle
$\angle ABD$ and $\angle ACD$ both intercept $\widehat{AD}$. $m\angle ABD = (4x - 7)^\circ$ and $m\angle ACD = (2x + 15)^\circ$. Find $m\widehat{AD}$.
- Both are half of $\widehat{AD}$, so they are equal: $4x - 7 = 2x + 15$, and $x = 11$.
- Each angle is $37^\circ$, so $m\widehat{AD} = 74^\circ$.
Answer: $74^\circ$
Worked example: A triangle in a semicircle
$\overline{AB}$ is a diameter and $C$ is on the circle with $m\angle CAB = 34^\circ$. Find $m\angle CBA$.
- $\angle ACB$ intercepts a semicircle, so it is $90^\circ$.
- $m\angle CBA = 180 - 90 - 34 = 56^\circ$.
Answer: $56^\circ$
Common mistake
Setting an inscribed angle equal to its arc, as if it were a central angle. Only an angle with its vertex at the center equals its arc; with the vertex on the circle, it is half. Check: find the vertex first.