Inscribed Angle Theorem - Statement, Proof, and Examples
Inscribed Angle Theorem - Statement, Proof, and Examples
TL;DR
The inscribed angle theorem states that an angle inscribed in a circle is half the central angle that subtends the same arc — so ∠inscribed = \tfrac{1}{2} ∠central. This article gives the exact statement, proves it in three cases, derives its main corollaries (angle in a semicircle is 90°; angles on the same arc are equal), and works through examples.
Why Every Angle Drawn to a Semicircle Comes Out to Exactly a Right Angle
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 ∠ABC is inscribed and ∠AOC is the central angle on the same arc AC (with O the centre), then:
∠ABC = \tfrac{1}{2} ∠AOC
Equivalently, the inscribed angle equals half its intercepted arc.
The Exact Statement
Fix a circle with centre O and an arc AC. Two angles "look at" that arc:
The central angle ∠AOC, with vertex at the centre O.
An inscribed 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:
∠ABC = \tfrac{1}{2} ∠AOC
As long as two angles subtend the identical arc, the one at the centre is exactly twice the one on the circle.
Proving The Theorem In Three Cases
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: ∠OAB = ∠OBA = θ. The central angle ∠AOC is the exterior angle of triangle OAB at O, so:
∠AOC = ∠OAB + ∠OBA = θ + θ = 2θ
Thus, ∠ABC = θ and ∠ABC = \tfrac{1}{2} ∠AOC.
Case 2 - the centre lies inside the angle. Draw the diameter BD through B, splitting ∠ABC into ∠ABD and ∠DBC. Case 1 applies to each half:
∠ABD = \tfrac{1}{2} ∠AOD, ∠DBC = \tfrac{1}{2} ∠DOC
Adding the two:
∠ABC = \tfrac{1}{2}(∠AOD + ∠DOC) = \tfrac{1}{2} ∠AOC.
Case 3 - the centre lies outside the angle. Now ∠ABC is the difference of two Case-1 angles:
∠ABC = ∠DBC - ∠DBA = \tfrac{1}{2} ∠DOC - \tfrac{1}{2} ∠DOA = \tfrac{1}{2} ∠AOC.
In all three positions the result is identical, which is why the theorem holds.
Two corollaries worth memorising
Angle in a semicircle is 90° (Thales' theorem). If AC is a diameter, the inscribed angle on the same arc is 90°.
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.
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:
∠inscribed = \tfrac{1}{2} × 80° = 40°.
Example 2
A student sees an inscribed angle of 35° and reports the central angle on the same arc as 35°. Spot the error.
The theorem says the central angle is twice the inscribed angle:
∠central = 2 × 35° = 70°.
Example 3
In a circle, AC is a diameter and B is any other point on the circle. Find ∠ABC.
Since AC is a diameter, the central angle ∠AOC = 180°. The inscribed angle ∠ABC is half of that:
∠ABC = \tfrac{1}{2} × 180° = 90°.
Example 4
Two inscribed angles, ∠ADB and ∠ACB, both subtend arc AB. If ∠ADB = 52°, find ∠ACB.
Both angles subtend the same arc, so:
∠ACB = ∠ADB = 52°.
Example 5
An inscribed angle intercepts an arc of 130°. Find the inscribed angle, then the central angle on that arc.
∠inscribed = \tfrac{1}{2} × 130° = 65°.
Example 6
A cyclic quadrilateral ABCD has ∠A = 95°. Find ∠C.
In a cyclic quadrilateral, opposite angles sum to 180°:
∠C = 180° - 95° = 85°.
Key Takeaways
The inscribed angle theorem says an inscribed angle is half the central angle.
Corollary: inscribed angles on the same arc are equal.