# Inscribed angle

In geometry, an inscribed angle is formed when two secant lines of a circle (or, in a degenerate case, when one secant line and one tangent line of that circle) intersect on the circle.

Typically it is easiest to think of an inscribed angle as being defined created by two chords of the circle sharing an endpoint.

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## Property

An inscribed angle is said to intercept an arc on the circle. The intercepted arc is the portion of the circle that is in the interior of the angle. The measure of the intercepted arc (equal to its central angle) is exactly twice the measure of the inscribed angle.

This single property has a number of consequences within the circle. For example, it allows one to prove that when two chords intersect in a circle, the products of the lengths of their pieces are equal. It also allows one to prove that the opposite angles of a cyclic quadrilateral are supplementary.

## Proof

To understand this proof, it is useful to draw a diagram.

### 1. Inscribed angles where one chord is a diameter

Missing image
InscribedAngle_1ChordDiam.png

Let O be the center of a circle. 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.

Angle BOA is a central angle; call it θ. Draw line OA. 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. They 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

[itex] 2 \psi + 180^\circ - \theta = 180^\circ [itex].

Subtract 180° from both sides,

[itex] 2 \psi = \theta [itex],

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

### 2. Inscribed angles with the center of the circle in their interior

Missing image
InscribedAngle_CenterCircle.png

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

[itex] \angle DVC = \angle DVE + \angle EVC [itex].

then let

[itex] \psi_0 = \angle DVC, [itex]
[itex] \psi_1 = \angle DVE, [itex]
[itex] \psi_2 = \angle EVC, [itex]

so that

[itex] \psi_0 = \psi_1 + \psi_2 \qquad \qquad (1) [itex]

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

[itex] \angle DOC = \angle DOE + \angle EOC. [itex]

Let

[itex] \theta_0 = \angle DOC, [itex]
[itex] \theta_1 = \angle DOE, [itex]
[itex] \theta_2 = \angle EOC, [itex]

so that

[itex] \theta_0 = \theta_1 + \theta_2 \qquad \qquad (2) [itex]

From Part One we know that [itex] \theta_1 = 2 \psi_1 [itex] and that [itex] \theta_2 = 2 \psi_2 [itex]. Combining these results with equation (2) yields

[itex] \theta_0 = 2 \psi_1 + 2 \psi_2 [itex]

therefore, by equation (1),

[itex] \theta_0 = 2 \psi_0. [itex]

### 3. Inscribed angles with the center of the circle in their exterior

Missing image
InscribedAngle_CenterCircleExt.png

[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 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

[itex] \angle DVC = \angle EVC - \angle DVE [itex].

then let

[itex] \psi_0 = \angle DVC, [itex]
[itex] \psi_1 = \angle DVE, [itex]
[itex] \psi_2 = \angle EVC, [itex]

so that

[itex] \psi_0 = \psi_2 - \psi_1 \qquad \qquad (3) [itex]

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

[itex] \angle DOC = \angle EOC - \angle DOE. [itex]

Let

[itex] \theta_0 = \angle DOC, [itex]
[itex] \theta_1 = \angle DOE, [itex]
[itex] \theta_2 = \angle EOC, [itex]

so that

[itex] \theta_0 = \theta_2 - \theta_1 \qquad \qquad (4) [itex]

From Part One we know that [itex] \theta_1 = 2 \psi_1 [itex] and that [itex] \theta_2 = 2 \psi_2 [itex]. Combining these results with equation (4) yields

[itex] \theta_0 = 2 \psi_2 - 2 \psi_1 [itex]

therefore, by equation (3),

[itex] \theta_0 = 2 \psi_0. [itex]

### 4. Conclusion

The measure of any inscribed angle is half the measure of its intercepted arc.

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