Step onto the circle and the angle you see is exactly half the one seen from the center.
The central angle was the view from the bull's-eye. Now walk out to the rim and look back at the same arc: the angle you see is exactly half. That one fact — the Inscribed Angle Theorem — is the beating heart of circle geometry. From it tumble a string of beautiful consequences: every spot on an arc sees its far arc at the same angle; the angle that looks across a diameter is a perfect right angle (Thales' theorem); and in any four-cornered figure drawn on a circle, opposite angles add to a straight line. This is the lesson the rest of the stage leans on.
We met the central angle in 18.2: its vertex sits at the center O, its sides are radii, and its measure equals the measure of the arc it cuts. Here the vertex moves. Slide it all the way out onto the circle itself, and a new kind of angle appears — the inscribed angle — with a tidy, surprising relationship to the central one. Everything in this lesson is one theorem and its four faces.
An inscribed angle has its vertex on the circle and two chords for its sides. The arc it subtends — the arc it "opens onto" — is the one cut off between those two chords, the arc that does not contain the vertex.
Hold the two pictures side by side. A central angle ∠AOC sits at the center O, arms along two radii. An inscribed angle ∠ABC sits out on the rim at B, arms along two chords. Both "look at" the same arc AC — but from very different seats.
An inscribed angle: vertex on the circle, both sides are chords. It subtends the arc trapped between those chords — the arc away from the vertex. Move the vertex to the center and you'd have a central angle instead.
Now drag the vertex B around the upper arc in the figure below. Watch the inscribed angle as B travels — and watch the central angle on the same arc stay put.
Here is the theorem, the one to memorize:
For the same arc AC, the inscribed angle is half the central angle:
∠ABC = 12 ∠AOC = 12 · (arc AC).
Why is it true? Take the cleanest case first — when one side of the inscribed angle runs straight through the center, so B, O, and A line up. Draw the radius OC. Now △OBC has OB = OC = r (both radii), so it is isosceles, and its two base angles are equal: ∠OBC = ∠OCB. Call each one x — that x is precisely our inscribed angle ∠ABC.
The central angle ∠AOC is the exterior angle of △OBC at O, and an exterior angle equals the sum of the two remote interior angles (a fact from triangle geometry). Those two remote angles are both x, so
∠AOC = x + x = 2x = 2 · ∠ABC. Divide by two: ∠ABC = 12 ∠AOC.
An arc AC measures 100°. Then the central angle ∠AOC is also 100°, and every inscribed angle that stands on that arc measures ½ · 100° = 50° — no matter where its vertex sits on the rest of the circle.
The half belongs to angles that share the same arc. An inscribed angle and a central angle are in a ½-to-1 ratio only when they look at the same arc. Mix up the arc and the relationship breaks.
Look again at what just happened in the first widget: as B slid around the upper arc, the inscribed angle never budged from 60°. That is no accident. Every inscribed angle on a given arc is half of the one central angle for that arc — and half of one fixed thing is one fixed thing.
All inscribed angles that subtend the same arc are equal. Wherever you plant the vertex on the far arc, the angle you see onto chord AC is identical.
This is the surveyor's secret and the photographer's too: every seat along one arc frames the same chord at the same angle. (Run it backward and you get a deep idea: if two points on the same side of a segment "see" it at equal angles, all four points lie on one circle — a first taste of cyclic points, which power the next section.)
Push the theorem to a special case. Let the arc the inscribed angle stands on be a whole semicircle — which happens exactly when its endpoints A and C are the two ends of a diameter. Then the central angle ∠AOC is a straight angle, 180°, and the inscribed angle is
∠ABC = 12 · 180° = 90°.
Any angle inscribed in a semicircle is a right angle. If AC is a diameter and B is anywhere else on the circle, then ∠ABC = 90°. The converse is just as useful: if ∠ABC = 90°, then B lies on the circle whose diameter is AC.
This little gem does real work. Need to build a right angle? Draw any diameter and pick a third point on the circle. Hunting for where a tangent touches, or for the foot of a perpendicular? Thales' theorem locates it. We will lean on it again in 18.5.
A quadrilateral whose four vertices all sit on one circle is called cyclic (or inscribed in the circle). Cyclic quadrilaterals carry a beautiful balance:
In a cyclic quadrilateral ABCD, opposite angles are supplementary:
∠A + ∠C = 180° and ∠B + ∠D = 180°.
The reason falls straight out of the inscribed-angle theorem. The angle at A and the angle at C are inscribed angles that stand on the two arcs which together make the entire circle. Each angle is half its arc, so
∠A + ∠C = 12(arc BCD) + 12(arc DAB) = 12 · 360° = 180°.
A cyclic quadrilateral has ∠A = 95°. Its opposite angle is forced: ∠C = 180° − 95° = 85°. You do not need to know ∠B or ∠D to find it — opposite angles always pair to a straight line.
Move the corners B and D below and watch ∠A and ∠C trade size while their sum holds at a perfect 180°.
One theorem, four faces — every one of them is "inscribed = half":
| Fact | Statement |
|---|---|
| Inscribed Angle Theorem | ∠ABC = ½ ∠AOC = ½ · (arc AC) |
| Same arc | all inscribed angles on one arc are equal |
| Semicircle (Thales) | angle on a diameter = 90° |
| Cyclic quadrilateral | opposite angles sum to 180° |
An inscribed angle sits on the rim with two chords for sides; a central angle sits at the center with two radii. On the same arc, the inscribed angle is exactly half the central one. From that single fact: same-arc angles are equal (half of one fixed angle), an angle across a diameter is a right angle (half of 180°), and the opposite corners of a four-sided figure on a circle add to a straight line (half of 360°). Keep the colors in mind — amber for the central angle and the arc it measures, green for the inscribed angle and every "= 90°", "= 180°", "equal ✓" conclusion. Next, in 18.4, we step back and sort every point, line, and second circle by a single comparison with the radius.
A central angle is 80°. What is the inscribed angle that stands on the same arc?
Inscribed = ½ central = ½ · 80° = 40°.
An inscribed angle measures 35°. What is the central angle on the same arc?
Central = 2 × inscribed = 2 · 35° = 70°. (And the arc it subtends measures 70° too.)
An inscribed angle subtends a diameter (so it stands on a semicircle). How big is it?
A diameter means a 180° arc, so the inscribed angle is ½ · 180° = 90° — Thales' theorem: the angle in a semicircle is a right angle.
In cyclic quadrilateral ABCD, ∠A = 95°. Find ∠C.
Opposite angles of a cyclic quadrilateral are supplementary, so ∠C = 180° − 95° = 85°.
Two inscribed angles look at the same arc, and one of them is 50°. What is the other?
Also 50° — all inscribed angles on the same arc are equal, since each is half of the one central angle for that arc.
An inscribed angle measures 60°. What is the measure of the arc it subtends?
The arc has measure 2 × the inscribed angle = 2 · 60° = 120°. (Equivalently the central angle is 120°, and the inscribed angle is half of it.)
Six questions to lock it in. Tap the answer you think is right.
The big idea of this lesson is a single, far-reaching theorem: an inscribed angle is half the central angle that stands on the same arc (∠ABC = ½ ∠AOC). Everything else — equal angles on one arc, the right angle in a semicircle (Thales), the supplementary opposite angles of a cyclic quadrilateral — is a consequence you can reconstruct from that one line and the words "same arc." The clean proof rests on tools students already own: a triangle with two radii is isosceles, and a central angle is the exterior angle of that triangle (so it equals twice a base angle). Asking a student to re-derive Thales' theorem from "half of 180°" is a better check of understanding than asking them to recall it.
The two errors to head off: (1) dropping the ½ — writing inscribed = central instead of inscribed = half central; and (2) using the wrong arc — applying the theorem to two angles that actually stand on different arcs, where no relationship holds. A third slip is failing to notice that "the chord is a diameter" is the trigger for the right angle in a semicircle. Insist that students name the arc out loud before they compute.
Common Core. This lesson develops G-C.A.2 — "Identify and describe relationships among inscribed angles, radii, and chords; the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle" — including the inscribed-angle theorem, the semicircle right angle, and the inscribed (cyclic) quadrilateral. It builds directly on the central-angle and arc-measure work of 18.2 and feeds the tangent and inscribed-circle constructions of 18.5.