Inscribed angle

For fixed points A and B, the set of points M in the plane for which the angle AMB is equal to α is an arc of a circle. The measure of ∠ AOB, where O is the center of the circle, is 2α.

The inscribed angle theorem states that an angle θ inscribed in a circle is half of the central angle 2θ that subtends the same arc on the circle. Therefore, the angle does not change as its vertex is moved to different positions on the circle.

Case: One chord is a diameter

Let O be the center of a circle, as in the diagram at right. Choose two points on the circle, and call them V and A. Draw line VO and extended past O so that it intersects the circle at point B which is diametrically opposite the point V. Draw an angle whose vertex is point V and whose sides pass through points A and B.

Draw line OA. Angle BOA is a central angle; call it θ. Lines OV and OA are both radii of the circle, so they have equal lengths. Therefore, triangle VOA is isosceles, so angle BVA (the inscribed angle) and angle VAO are equal; let each of them be denoted as ψ.

Angles BOA and AOV add up to 180°, since line VB passing through O is a straight line. Therefore, angle AOV measures 180° − θ.

It is known that the three angles of a triangle add up to 180°, and the three angles of triangle VOA are:

180° − θ

ψ

ψ.

Therefore,

${\displaystyle 2\psi +180^{\circ }-\theta =180^{\circ }.}$

Subtract

${\displaystyle (180^{\circ }-\theta )}$

from both sides,

${\displaystyle 2\psi =\theta ,}$

where θ is the central angle subtending arc AB and ψ is the inscribed angle subtending arc AB.

Case: Center interior to angle

Given a circle whose center is point O, choose three points V, C, and D on the circle. Draw lines VC and VD: angle DVC is an inscribed angle. Now draw line VO and extend it past point O so that it intersects the circle at point E. Angle DVC subtends arc DC on the circle.

Suppose this arc includes point E within it. Point E is diametrically opposite to point V. Angles DVE and EVC are also inscribed angles, but both of these angles have one side which passes through the center of the circle, therefore the theorem from the above Part 1 can be applied to them.

Therefore,

${\displaystyle \angle DVC=\angle DVE+\angle EVC.}$

then let

${\displaystyle \psi _{0}=\angle DVC,}$

${\displaystyle \psi _{1}=\angle DVE,}$

${\displaystyle \psi _{2}=\angle EVC,}$

so that

${\displaystyle \psi _{0}=\psi _{1}+\psi _{2}.\qquad \qquad (1)}$

Draw lines OC and OD. Angle DOC is a central angle, but so are angles DOE and EOC, and

${\displaystyle \angle DOC=\angle DOE+\angle EOC.}$

Let

${\displaystyle \theta _{0}=\angle DOC,}$

${\displaystyle \theta _{1}=\angle DOE,}$

${\displaystyle \theta _{2}=\angle EOC,}$

so that

${\displaystyle \theta _{0}=\theta _{1}+\theta _{2}.\qquad \qquad (2)}$

From Part One we know that ${\displaystyle \theta _{1}=2\psi _{1}}$ and that ${\displaystyle \theta _{2}=2\psi _{2}}$. Combining these results with equation (2) yields

${\displaystyle \theta _{0}=2\psi _{1}+2\psi _{2}=2(\psi _{1}+\psi _{2})}$

therefore, by equation (1),

${\displaystyle \theta _{0}=2\psi _{0}.}$

Case: Center exterior to angle

The previous case can be extended to cover the case where the measure of the inscribed angle is the difference between two inscribed angles as discussed in the first part of this proof.

Given a circle whose center is point O, choose three points V, C, and D on the circle. Draw lines VC and VD: angle DVC is an inscribed angle. Now draw line VO and extend it past point O so that it intersects the circle at point E. Angle DVC subtends arc DC on the circle.

Suppose this arc does not include point E within it. Point E is diametrically opposite to point V. Angles EVD and EVC are also inscribed angles, but both of these angles have one side which passes through the center of the circle, therefore the theorem from the above Part 1 can be applied to them.

Therefore,

${\displaystyle \angle DVC=\angle EVC-\angle EVD}$.

then let

${\displaystyle \psi _{0}=\angle DVC,}$

${\displaystyle \psi _{1}=\angle EVD,}$

${\displaystyle \psi _{2}=\angle EVC,}$

so that

${\displaystyle \psi _{0}=\psi _{2}-\psi _{1}.\qquad \qquad (3)}$

Draw lines OC and OD. Angle DOC is a central angle, but so are angles EOD and EOC, and

${\displaystyle \angle DOC=\angle EOC-\angle EOD.}$

Let

${\displaystyle \theta _{0}=\angle DOC,}$

${\displaystyle \theta _{1}=\angle EOD,}$

${\displaystyle \theta _{2}=\angle EOC,}$

so that

${\displaystyle \theta _{0}=\theta _{2}-\theta _{1}.\qquad \qquad (4)}$

From Part One we know that ${\displaystyle \theta _{1}=2\psi _{1}}$ and that ${\displaystyle \theta _{2}=2\psi _{2}}$. Combining these results with equation (4) yields

${\displaystyle \theta _{0}=2\psi _{2}-2\psi _{1}}$

therefore, by equation (3),

${\displaystyle \theta _{0}=2\psi _{0}.}$

Animated gif of proof of the inscribed angle theorem. The large triangle that is inscribed in the circle gets subdivided into three smaller triangles, all of which are isosceles because their upper two sides are radii of the circle. Inside each isosceles triangle the pair of base angles are equal to each other, and are half of 180° minus the apex angle at the circle's center. Adding up these isosceles base angles yields the theorem, namely that the inscribed angle, ${\displaystyle \psi }$, is half the central angle, ${\displaystyle \theta }$.

By a similar argument, the angle between a chord and the tangent line at one of its intersection points equals half of the central angle subtended by the chord. See also Tangent lines to circles.