Editing Inscribed angle (section)

==Theorem==

===Statement===

[[File:ArcCapable.gif|thumb|class=skin-invert-image|For fixed points {{mvar|A}} and {{mvar|B}}, the set of points ''M'' in the plane, for which the angle {{math|∠''AMB''}} is equal to&nbsp;''α'', is an arc of a circle. The measure of {{math|∠''AOB''}}, where {{mvar|O}} is the center of the circle, is&nbsp;{{math|2''α''}}.]]

The inscribed angle theorem states that an angle {{mvar|θ}} inscribed in a circle is half of the central angle {{math|2''θ''}} that [[intercepted arc|intercepts]] the same [[arc (geometry)|arc]] on the circle.  Therefore, the angle does not change as its [[vertex (geometry)|vertex]] is moved to different positions on the same arc of the circle.

===Proof===

====Inscribed angles where one chord is a diameter====
[[File:InscribedAngle 1ChordDiam.svg|thumb|class=skin-invert-image|Case: One chord is a diameter]]
Let {{mvar|O}} be the center of a circle, as in the diagram at right.  Choose two points on the circle, and call them {{mvar|V}} and {{mvar|A}}.  Designate point {{mvar|B}} to be [[diametrically opposite]] point {{mvar|V}}.  Draw chord {{Mvar|VB}}, a diameter containing point {{Mvar|O}}.  Draw chord {{Mvar|VA}}.  Angle {{Math|∠''BVA''}} is an inscribed angle that intercepts arc {{Mvar|{{overarc|AB}}}}; denote it as {{Mvar|ψ}}.  Draw line {{mvar|OA}}.  Angle {{math|∠''BOA''}} is a [[central angle]] that also intercepts arc {{Mvar|{{overarc|AB}}}}; denote it as {{mvar|θ}}.

Lines {{mvar|OV}} and {{mvar|OA}} are both [[radius|radii]] of the circle, so they have equal lengths.  Therefore, triangle {{math|△''VOA''}} is [[isosceles]], so angle {{math|∠''BVA''}} and angle {{math|∠''VAO''}} are equal.

Angles {{math|∠''BOA''}} and {{math|∠''AOV''}} are [[supplementary angle|supplementary]], summing to a [[straight angle]] (180°), so angle {{math|∠''AOV''}} measures {{math|180° &minus; ''θ''}}.

The three angles of triangle {{math|△''VOA''}} [[sum of angles of a triangle|must sum to {{math|180°}}]]:

<math display=block>(180^\circ - \theta) + \psi + \psi = 180^\circ.</math>

Adding <math>\theta - 180^\circ</math> to both sides yields

<math display=block>2\psi = \theta.</math>

====Inscribed angles with the center of the circle in their interior====
[[File:Circle-angles-21add-inscribed.svg|thumb|class=skin-invert-image|
Case: Center interior to angle
{{legend-line|solid blue|{{math|1=''&psi;''{{sub|0}} = ∠''DVC'', ''&theta;''{{sub|0}} = ∠''DOC''}}}}
{{legend-line|solid green|{{math|1=''&psi;''{{sub|1}} = ∠''EVD'', ''&theta;''{{sub|1}} = ∠''EOD''}}}}
{{legend-line|solid red|{{math|1=''&psi;''{{sub|2}} = ∠''EVC'', ''&theta;''{{sub|2}} = ∠''EOC''}}}}
]]
Given a circle whose center is point {{mvar|O}}, choose three points {{mvar|V, C, D}} on the circle.  Draw lines {{mvar|VC}} and {{mvar|VD}}: angle {{math|∠''DVC''}} is an inscribed angle.  Now draw line {{mvar|OV}} and extend it past point {{mvar|O}} so that it intersects the circle at point {{mvar|E}}.  Angle {{math|∠''DVC''}} intercepts arc {{mvar|{{overarc|DC}}}} on the circle.

Suppose this arc includes point {{mvar|E}} within it.  Point {{mvar|E}} is diametrically opposite to point {{mvar|V}}.  Angles {{math|∠''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,

<math display=block> \angle DVC = \angle DVE + \angle EVC. </math>

then let

<math display=block>\begin{align}
  \psi_0 &= \angle DVC, \\
  \psi_1 &= \angle DVE, \\
  \psi_2 &= \angle EVC, 
\end{align}</math>

so that

<math display=block> \psi_0 = \psi_1 + \psi_2. \qquad \qquad (1) </math>

Draw lines {{mvar|OC}} and {{mvar|OD}}.  Angle {{math|∠''DOC''}} is a central angle, but so are angles {{math|∠''DOE''}} and {{math|∠''EOC''}}, and
<math display=block> \angle DOC = \angle DOE + \angle EOC. </math>

Let

<math display=block>\begin{align}
  \theta_0 &= \angle DOC, \\
  \theta_1 &= \angle DOE, \\
  \theta_2 &= \angle EOC, 
\end{align}</math>

so that

<math display=block> \theta_0 = \theta_1 + \theta_2. \qquad \qquad (2) </math>

From Part One we know that <math> \theta_1 = 2 \psi_1 </math> and that <math> \theta_2 = 2 \psi_2 </math>.  Combining these results with equation (2) yields

<math display=block> \theta_0 = 2 \psi_1 + 2 \psi_2 = 2(\psi_1 + \psi_2) </math>

therefore, by equation (1),

<math display=block> \theta_0 = 2 \psi_0. </math>

====Inscribed angles with the center of the circle in their exterior====
[[Image:InscribedAngle CenterCircleExtV2.svg|thumb|Case: Center exterior to angle
{{legend-line|solid orange|{{math|1=''&psi;''{{sub|0}} = ∠''DVC'', ''&theta;''{{sub|0}} = ∠''DOC''}}}}
{{legend-line|solid green|{{math|1=''&psi;''{{sub|1}} = ∠''EVD'', ''&theta;''{{sub|1}} = ∠''EOD''}}}}
{{legend-line|solid blue|{{math|1=''&psi;''{{sub|2}} = ∠''EVC'', ''&theta;''{{sub|2}} = ∠''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 {{mvar|O}}, choose three points {{mvar|V, C, D}} on the circle.  Draw lines {{mvar|VC}} and {{mvar|VD}}: angle {{math|∠''DVC''}} is an inscribed angle.  Now draw line {{mvar|OV}} and extend it past point {{mvar|O}} so that it intersects the circle at point {{mvar|E}}.  Angle {{math|∠''DVC''}} intercepts arc {{mvar|{{overarc|DC}}}} on the circle.

Suppose this arc does not include point {{mvar|E}} within it.  Point {{mvar|E}} is diametrically opposite to point {{mvar|V}}.  Angles {{math|∠''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,

<math display=block> \angle DVC = \angle EVC - \angle EVD. </math>

then let

<math display=block>\begin{align}
  \psi_0 &= \angle DVC, \\
  \psi_1 &= \angle EVD, \\
  \psi_2 &= \angle EVC, 
\end{align}</math>

so that

<math display=block> \psi_0 = \psi_2 - \psi_1. \qquad \qquad (3) </math>

Draw lines {{mvar|OC}} and {{mvar|OD}}.  Angle {{math|∠''DOC''}} is a central angle, but so are angles {{math|∠''EOD''}} and {{math|∠''EOC''}}, and

<math display=block> \angle DOC = \angle EOC - \angle EOD. </math>

Let

<math display=block>\begin{align}
  \theta_0 &= \angle DOC, \\
  \theta_1 &= \angle EOD, \\
  \theta_2 &= \angle EOC, 
\end{align}</math>

so that

<math display=block> \theta_0 = \theta_2 - \theta_1. \qquad \qquad (4) </math>

From Part One we know that <math> \theta_1 = 2 \psi_1 </math> and that <math> \theta_2 = 2 \psi_2 </math>.  Combining these results with equation (4) yields
<math display=block> \theta_0 = 2 \psi_2 - 2 \psi_1 </math>
therefore, by equation (3),
<math display=block> \theta_0 = 2 \psi_0. </math>


[[File:Animated gif of proof of the inscribed angle theorem.gif|thumb|class=skin-invert-image|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, {{mvar|&psi;}}, is half the central angle, {{mvar|&theta;}}.]]

===Corollary===
[[Image:Inscribed angle theorem4.svg|thumb|The angle {{mvar|θ}} between a chord and a tangent is half the arc belonging to the chord.]]

By a similar argument, the angle between a [[Chord (geometry)|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]].