Inscribed angle

The inscribed angle theorem, sometimes called the central angle theorem, relates the measure of an inscribed angle to that of the central angle intercepting the same arc. Its oldest appearance is in Propositions 20–21 in Book 3 of Euclid's Elements.

Theorem

Statement

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

Proof

Inscribed angles where 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. Designate point B to be diametrically opposite point V. Draw chord VB, a diameter containing point O. Draw chord VA. Angle ∠BVA is an inscribed angle that intercepts arc AB; denote it as ψ. Draw line OA. Angle ∠BOA is a central angle that also intercepts arc AB; denote it as θ.

Lines OV and OA are both radii of the circle, so they have equal lengths. Therefore, triangle △VOA is isosceles, so angle ∠BVA and angle ∠VAO are equal.

Angles ∠BOA and ∠AOV are supplementary, summing to a straight angle (180°), so angle ∠AOV measures 180° − θ.

The three angles of triangle △VOA must sum to 180°:

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

Adding ${\displaystyle \theta -{180}^{\circ }}$ to both sides yields

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

Inscribed angles with the center of the circle in their interior

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

Suppose this arc includes point E within it. Point E is diametrically opposite to point V. Angles ∠DVE, ∠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,

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

then let

$${\displaystyle \begin{array}{cc}{\psi }_0& =\mathrm{\angle }DVC,\\ {\psi }_1& =\mathrm{\angle }DVE,\\ {\psi }_2& =\mathrm{\angle }EVC,\end{array}}$$

so that

$${\displaystyle {\psi }_0={\psi }_1+{\psi }_2.(1)}$$

Draw lines OC and OD. Angle ∠DOC is a central angle, but so are angles ∠DOE and ∠EOC, and $${\displaystyle \mathrm{\angle }DOC=\mathrm{\angle }DOE+\mathrm{\angle }EOC.}$$

Let

$${\displaystyle \begin{array}{cc}{\theta }_0& =\mathrm{\angle }DOC,\\ {\theta }_1& =\mathrm{\angle }DOE,\\ {\theta }_2& =\mathrm{\angle }EOC,\end{array}}$$

so that

$${\displaystyle {\theta }_0={\theta }_1+{\theta }_2.(2)}$$

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),

$${\displaystyle {\theta }_0=2{\psi }_0.}$$

Inscribed angles with the center of the circle in their exterior

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, D on the circle. Draw lines VC and VD: angle ∠DVC is an inscribed angle. Now draw line OV and extend it past point O so that it intersects the circle at point E. Angle ∠DVC intercepts 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, ∠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,

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

then let

$${\displaystyle \begin{array}{cc}{\psi }_0& =\mathrm{\angle }DVC,\\ {\psi }_1& =\mathrm{\angle }EVD,\\ {\psi }_2& =\mathrm{\angle }EVC,\end{array}}$$

so that

$${\displaystyle {\psi }_0={\psi }_2-{\psi }_1.(3)}$$

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

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

Let

$${\displaystyle \begin{array}{cc}{\theta }_0& =\mathrm{\angle }DOC,\\ {\theta }_1& =\mathrm{\angle }EOD,\\ {\theta }_2& =\mathrm{\angle }EOC,\end{array}}$$

so that

$${\displaystyle {\theta }_0={\theta }_2-{\theta }_1.(4)}$$

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), $${\displaystyle {\theta }_0=2{\psi }_0.}$$

Corollary

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.