Why Every Angle Drawn to a Semicircle Comes Out to Exactly a Right Angle
Pick any point on a circle, join it to the two ends of a diameter, and the angle you make is always exactly $90°$ - no matter which point you chose. That surprising, always-true fact is one line of a single deeper theorem about circles.
The inscribed angle theorem states that the measure of an inscribed angle (its vertex on the circle, its two sides being chords) is half the measure of the central angle that subtends the same arc. In symbols, if $\angle ABC$ is inscribed and $\angle AOC$ is the central angle on the same arc $AC$ (with $O$ the centre), then:
$$\angle ABC = \frac{1}{2},\angle AOC$$
Equivalently, the inscribed angle equals half its intercepted arc. This theorem builds on the central angle in geometry and on chords, and it is one of the most-used facts about circles.
By the end you will be able to state the theorem, prove it for every position of the vertex, and use its two famous corollaries.
The Exact Statement
Fix a circle with centre $O$ and an arc $AC$. Two angles "look at" that arc:
The central angle $\angle AOC$, with vertex at the centre $O$.
An inscribed angle $\angle ABC$, with vertex $B$ on the circle and sides that are the chords $BA$ and $BC$.
The theorem says the inscribed angle is always half the central angle on the same arc:
$$\angle ABC = \frac{1}{2},\angle AOC$$
The crucial words are same arc. As long as two angles subtend the identical arc, the one at the centre is exactly twice the one on the circle, however $B$ is positioned along the rest of the circle. The chords $BA$ and $BC$ are ordinary chords of a circle.
Proving The Theorem In Three Cases
The vertex $B$ can sit in three positions relative to the diameter through $B$. Each case reduces to the same isosceles-triangle idea, using the fact that $OA = OB = OC$ are all radii.
Case 1 - the centre lies on one side of the angle. Suppose side $BC$ passes through the centre $O$, so $BC$ is a diameter. Triangle $OAB$ is isosceles ($OA = OB$), so its base angles are equal: $\angle OAB = \angle OBA = \theta$. The central angle $\angle AOC$ is the exterior angle of triangle $OAB$ at $O$, and an exterior angle equals the sum of the two remote interior angles:
$$\angle AOC = \angle OAB + \angle OBA = \theta + \theta = 2\theta$$
Since the inscribed angle $\angle ABC = \theta$, we get $\angle ABC = \tfrac{1}{2}\angle AOC$.
Case 2 - the centre lies inside the angle. Draw the diameter $BD$ through $B$, splitting $\angle ABC$ into $\angle ABD$ and $\angle DBC$. Case 1 applies to each half:
$$\angle ABD = \tfrac{1}{2}\angle AOD, \qquad \angle DBC = \tfrac{1}{2}\angle DOC$$
Adding the two:
$$\angle ABC = \tfrac{1}{2}(\angle AOD + \angle DOC) = \tfrac{1}{2}\angle AOC$$
Case 3 - the centre lies outside the angle. Draw the diameter $BD$ again; now $\angle ABC$ is the difference of two Case-1 angles rather than the sum:
$$\angle ABC = \angle DBC - \angle DBA = \tfrac{1}{2}\angle DOC - \tfrac{1}{2}\angle DOA = \tfrac{1}{2}\angle AOC$$
In all three positions the result is identical, which is why the theorem holds for every inscribed angle.
Two corollaries worth memorising
The theorem's power shows in two consequences that appear constantly in problems.
Angle in a semicircle is $90°$ (Thales' theorem). If $AC$ is a diameter, the central angle $\angle AOC$ is a straight angle, $180°$. Any inscribed angle on the same arc is half of that: $\tfrac{1}{2}\times 180° = 90°$. So every triangle drawn inside a semicircle with the diameter as one side has a right angle at the third vertex.
Angles on the same arc are equal. Two inscribed angles that subtend the same arc both equal half of the one central angle, so they are equal to each other. Move the vertex anywhere along the arc and the angle does not change.
A third, related result: the opposite angles of a cyclic quadrilateral (four vertices on one circle) sum to $180°$, which also follows from the theorem applied to the two arcs.
Examples of Inscribed Angle Theorem
Example 1
A central angle subtends an arc of $80°$. Find the inscribed angle on the same arc.
The inscribed angle is half the central angle:
$$\angle \text{inscribed} = \tfrac{1}{2}\times 80° = 40°$$
The inscribed angle is $40°$.
Example 2
A student sees an inscribed angle of $35°$ and reports the central angle on the same arc as $35°$ too. Spot the error.
A natural first move is to assume the two angles looking at one arc are equal. But the theorem says the central angle is twice the inscribed angle, not equal to it, so copying the value skips the factor of $2$.
Double the inscribed angle:
$$\angle \text{central} = 2 \times 35° = 70°$$
The central angle is $70°$. The inscribed angle is the smaller one; the central angle is always double.
Example 3
In a circle, $AC$ is a diameter and $B$ is any other point on the circle. Find $\angle ABC$.
Since $AC$ is a diameter, the central angle $\angle AOC = 180°$. The inscribed angle $\angle ABC$ on the same arc is half of that:
$$\angle ABC = \tfrac{1}{2}\times 180° = 90°$$
The angle is $90°$, a right angle, for every position of $B$ - this is Thales' theorem.
Example 4
Two inscribed angles, $\angle ADB$ and $\angle ACB$, both subtend arc $AB$. If $\angle ADB = 52°$, find $\angle ACB$.
Both angles subtend the same arc $AB$, so by the same-arc corollary they are equal:
$$\angle ACB = \angle ADB = 52°$$
The angle $\angle ACB$ is $52°$.
Example 5
An inscribed angle intercepts an arc of $130°$. Find the inscribed angle, then the central angle on that arc.
An inscribed angle equals half its intercepted arc:
$$\angle \text{inscribed} = \tfrac{1}{2}\times 130° = 65°$$
The central angle equals the full arc:
$$\angle \text{central} = 130°$$
The inscribed angle is $65°$ and the central angle is $130°$, confirming inscribed $= \tfrac{1}{2}$ central.
Example 6
A cyclic quadrilateral $ABCD$ has $\angle A = 95°$. Find $\angle C$.
In a cyclic quadrilateral, opposite angles are supplementary, a consequence of the inscribed angle theorem applied to the two arcs $\angle A$ and $\angle C$ subtend:
$$\angle A + \angle C = 180°$$
$$\angle C = 180° - 95° = 85°$$
The opposite angle $\angle C$ is $85°$. This is the tool bearing designers use to check that four points lie on one circle.
Where the Inscribed Angle Theorem Earns its Keep: Fixing a Right Angle From a Circle
The theorem matters because it manufactures a guaranteed right angle and equal angles from nothing but a circle, which turns up in tools, construction, and design.
Finding a right angle without a set-square. Thales' theorem lets a builder mark a perfect $90°$ corner using only a circle and a diameter, a trick used since antiquity when no square was at hand.
Testing whether points lie on a circle. Since opposite angles of a cyclic quadrilateral sum to $180°$, a designer can check four holes or joints are concyclic just by measuring two angles, useful in mechanical linkages and gear layouts.
Why half, and why it is constant. The destination is a stable angle. Because the inscribed angle depends only on the arc and not on where the vertex sits, a viewing angle stays the same all along an arc — the geometry behind why a whole row of seats can share one sightline to a stage.
Thales' construction of a right angle in a semicircle is one of the oldest recorded uses, credited to Thales of Miletus and still the standard way to raise a perpendicular with compass and straightedge.
Mistakes to watch for
Mistake 1: Treating the inscribed and central angles as equal
Where it slips in: Reading a problem that gives one angle on an arc and asks for the other.
Don't do this: Copy the inscribed angle straight across as the central angle (or vice versa).
The correct way: The central angle is twice the inscribed angle on the same arc; the inscribed angle is half the central. Decide which one you have before applying the factor of $2$. The rusher who assumes "same arc means same angle" loses the factor every time.
Mistake 2: Confusing the intercepted arc with a different arc
Where it slips in: Applying the theorem when the vertex looks toward the wrong side of the circle.
Don't do this: Use the arc on the near side of the vertex instead of the arc the chords actually cut off.
The correct way: The intercepted arc is the one between the two chords, on the far side from the vertex. Trace both chords to the circle and identify the arc they enclose before halving it.
Mistake 3: Forgetting the diameter condition for the $90°$ result
Where it slips in: Claiming an inscribed angle is $90°$ without checking that its chord is a diameter.
Don't do this: Assume any inscribed triangle has a right angle.
The correct way: The angle in a semicircle is $90°$ only when the subtended chord is a diameter (central angle $180°$). If the chord is shorter, the inscribed angle is less than $90°$. Confirm the chord passes through the centre first.
Key Takeaways
The inscribed angle theorem says an inscribed angle is half the central angle on the same arc.
Equivalently, an inscribed angle equals half its intercepted arc.
The proof splits into three cases (centre on a side, inside, outside), all resting on isosceles radii.
Corollary 1: an angle in a semicircle is $90°$ (Thales' theorem).
Corollary 2: inscribed angles on the same arc are equal, and opposite angles of a cyclic quadrilateral sum to $180°$.
A practical next step
Practice these problems to solidify your understanding. For each, first identify the intercepted arc, then decide whether you are halving a central angle or doubling an inscribed one.
A central angle is $110°$. Find the inscribed angle on the same arc. (Answer to Question 1: $\tfrac{1}{2}\times 110° = 55°$.)
An inscribed angle is $48°$. Find the central angle on the same arc. (Answer to Question 2: $2 \times 48° = 96°$.)
To work through circle theorems with a teacher, explore Bhanzu's geometry tutor, our high school math tutor sessions, or math classes online. To see a trainer prove the inscribed angle theorem live, you can book a free demo class.
Read More
Arc length - how the intercepted arc that the inscribed angle halves is measured.
Types of angles - where inscribed and central angles sit among angle types.
Parts of a circle - centre, radius, chord, arc, and the other elements a circle theorem uses.
Semicircle - the half-circle behind the $90°$ Thales corollary.
Segment of circle - the region a chord cuts off, related to inscribed-angle problems.
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