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Inscribed angle

From Wikipedia, the free encyclopedia
Angle formed in the interior of a circle
The inscribed angleθ circle.
  Inscribed angleθ onmajor arc
  Supplementary inscribed angleθ on minor arc

Ingeometry, aninscribed angle is theangle formed in the interior of acircle when twochords intersect on the circle. It can also be defined as the anglesubtended at a point on the circle by two given points on the circle.

Equivalently, an inscribed angle is defined by two chords of the circle sharing an endpoint.

Theinscribed angle theorem relates themeasure of an inscribed angle to that of thecentral angle intercepting the samearc.

The inscribed angle theorem appears as Proposition 20 in Book 3 ofEuclid'sElements.

Note that this theorem is not to be confused with theAngle bisector theorem, which also involves angle bisection (but of an angle of a triangle not inscribed in a circle).

Theorem

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Statement

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For fixed pointsA andB, the set of pointsM in the plane, for which the angle∠AMB is equal to α, is an arc of a circle. The measure of∠AOB, whereO is the center of the circle, is 2α.

The inscribed angle theorem states that an angleθ inscribed in a circle is half of the central angle2θ thatintercepts the samearc on the circle. Therefore, the angle does not change as itsvertex is moved to different positions on the same arc of the circle.

Proof

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Inscribed angles where one chord is a diameter

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Case: One chord is a diameter

LetO be the center of a circle, as in the diagram at right. Choose two points on the circle, and call themV andA. Designate pointB to bediametrically opposite pointV. Draw chordVB, a diameter containing pointO. Draw chordVA. Angle∠BVA is an inscribed angle that intercepts arcAB; denote it asψ. Draw lineOA. Angle∠BOA is acentral angle that also intercepts arcAB; denote it asθ.

LinesOV andOA are bothradii of the circle, so they have equal lengths. Therefore, triangle△VOA isisosceles, so angle∠BVA and angle∠VAO are equal.

Angles∠BOA and∠AOV aresupplementary, summing to astraight angle (180°), so angle∠AOV measures180° −θ.

The three angles of triangle△VOAmust sum to180°:

(180∘−θ)+ψ+ψ=180∘.{\displaystyle (180^{\circ }-\theta )+\psi +\psi =180^{\circ }.}

Addingθ−180∘{\displaystyle \theta -180^{\circ }} to both sides yields

2ψ=θ.{\displaystyle 2\psi =\theta .}

Inscribed angles with the center of the circle in their interior

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Case: Center interior to angle
  ψ0 = ∠DVC,θ0 = ∠DOC
  ψ1 = ∠EVD,θ1 = ∠EOD
  ψ2 = ∠EVC,θ2 = ∠EOC

Given a circle whose center is pointO, choose three pointsV, C, D on the circle. Draw linesVC andVD: angle∠DVC is an inscribed angle. Now draw lineOV and extend it past pointO so that it intersects the circle at pointE. Angle∠DVC intercepts arcDC on the circle.

Suppose this arc includes pointE within it. PointE is diametrically opposite to pointV. 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,

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

then let

ψ0=∠DVC,ψ1=∠DVE,ψ2=∠EVC,{\displaystyle {\begin{aligned}\psi _{0}&=\angle DVC,\\\psi _{1}&=\angle DVE,\\\psi _{2}&=\angle EVC,\end{aligned}}}

so that

ψ0=ψ1+ψ2.(1){\displaystyle \psi _{0}=\psi _{1}+\psi _{2}.\qquad \qquad (1)}

Draw linesOC andOD. Angle∠DOC is a central angle, but so are angles∠DOE and∠EOC, and∠DOC=∠DOE+∠EOC.{\displaystyle \angle DOC=\angle DOE+\angle EOC.}

Let

θ0=∠DOC,θ1=∠DOE,θ2=∠EOC,{\displaystyle {\begin{aligned}\theta _{0}&=\angle DOC,\\\theta _{1}&=\angle DOE,\\\theta _{2}&=\angle EOC,\end{aligned}}}

so that

θ0=θ1+θ2.(2){\displaystyle \theta _{0}=\theta _{1}+\theta _{2}.\qquad \qquad (2)}

From Part One we know thatθ1=2ψ1{\displaystyle \theta _{1}=2\psi _{1}} and thatθ2=2ψ2{\displaystyle \theta _{2}=2\psi _{2}}. Combining these results with equation (2) yields

θ0=2ψ1+2ψ2=2(ψ1+ψ2){\displaystyle \theta _{0}=2\psi _{1}+2\psi _{2}=2(\psi _{1}+\psi _{2})}

therefore, by equation (1),

θ0=2ψ0.{\displaystyle \theta _{0}=2\psi _{0}.}

Inscribed angles with the center of the circle in their exterior

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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 thedifference between two inscribed angles as discussed in the first part of this proof.

Given a circle whose center is pointO, choose three pointsV, C, D on the circle. Draw linesVC andVD: angle∠DVC is an inscribed angle. Now draw lineOV and extend it past pointO so that it intersects the circle at pointE. Angle∠DVC intercepts arcDC on the circle.

Suppose this arc does not include pointE within it. PointE is diametrically opposite to pointV. 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,

∠DVC=∠EVC−∠EVD.{\displaystyle \angle DVC=\angle EVC-\angle EVD.}

then let

ψ0=∠DVC,ψ1=∠EVD,ψ2=∠EVC,{\displaystyle {\begin{aligned}\psi _{0}&=\angle DVC,\\\psi _{1}&=\angle EVD,\\\psi _{2}&=\angle EVC,\end{aligned}}}

so that

ψ0=ψ2−ψ1.(3){\displaystyle \psi _{0}=\psi _{2}-\psi _{1}.\qquad \qquad (3)}

Draw linesOC andOD. Angle∠DOC is a central angle, but so are angles∠EOD and∠EOC, and

∠DOC=∠EOC−∠EOD.{\displaystyle \angle DOC=\angle EOC-\angle EOD.}

Let

θ0=∠DOC,θ1=∠EOD,θ2=∠EOC,{\displaystyle {\begin{aligned}\theta _{0}&=\angle DOC,\\\theta _{1}&=\angle EOD,\\\theta _{2}&=\angle EOC,\end{aligned}}}

so that

θ0=θ2−θ1.(4){\displaystyle \theta _{0}=\theta _{2}-\theta _{1}.\qquad \qquad (4)}

From Part One we know thatθ1=2ψ1{\displaystyle \theta _{1}=2\psi _{1}} and thatθ2=2ψ2{\displaystyle \theta _{2}=2\psi _{2}}. Combining these results with equation (4) yieldsθ0=2ψ2−2ψ1{\displaystyle \theta _{0}=2\psi _{2}-2\psi _{1}}therefore, by equation (3),θ0=2ψ0.{\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.

Corollary

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The angleθ between a chord and a tangent is half the arc belonging to the chord.

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

Applications

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Proof without words using the inscribed angle theorem that opposite angles of acyclic quadrilateral are supplementary:
2𝜃 + 2𝜙 = 360° ∴ 𝜃 + 𝜙 = 180°

The inscribed angletheorem is used in many proofs of elementaryEuclidean geometry of the plane. A special case of the theorem isThales's theorem, which states that the angle subtended by adiameter is always 90°, i.e., a right angle. As a consequence of the theorem, opposite angles ofcyclic quadrilaterals sum to 180°; conversely, any quadrilateral for which this is true can be inscribed in a circle. As another example, the inscribed angle theorem is the basis for several theorems related to thepower of a point with respect to a circle. Further, it allows one to prove that when two chords intersect in a circle, the products of the lengths of their pieces are equal.

Inscribed angle theorems for ellipses, hyperbolas and parabolas

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Inscribed angle theorems exist for ellipses, hyperbolas and parabolas too. The essential differences are the measurements of an angle. (An angle is considered a pair of intersecting lines.)

References

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External links

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