Editing Inscribed angle (section)

====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>