Inscribed angles where one chord is a diameter

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 OV 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, 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 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

Case: Center interior to angle

  φ₀ = ∠DVC, θ₀ = ∠DOC

  φ₁ = ∠EVD, θ₁ = ∠EOD

  φ₂ = ∠EVC, θ₂ = ∠EOC

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 subtends 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,

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

then let

$${\displaystyle {\begin{aligned}\psi _{0}&=\angle DVC,\\\psi _{1}&=\angle DVE,\\\psi _{2}&=\angle EVC,\end{aligned}}}$$

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

$${\displaystyle {\begin{aligned}\theta _{0}&=\angle DOC,\\\theta _{1}&=\angle DOE,\\\theta _{2}&=\angle EOC,\end{aligned}}}$$

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),

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

Inscribed angles with the center of the circle in their exterior

Case: Center exterior to angle

  ψ₀ = ∠DVC, θ₀ = ∠DOC

  ψ₁ = ∠EVD, θ₁ = ∠EOD

  ψ₂ = ∠EVC, θ₂ = ∠EOC

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 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, ∠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

$${\displaystyle {\begin{aligned}\psi _{0}&=\angle DVC,\\\psi _{1}&=\angle EVD,\\\psi _{2}&=\angle EVC,\end{aligned}}}$$

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

$${\displaystyle {\begin{aligned}\theta _{0}&=\angle DOC,\\\theta _{1}&=\angle EOD,\\\theta _{2}&=\angle EOC,\end{aligned}}}$$

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, ψ, is half the central angle, θ.