A circle is centered on point B. Points A, C
and D lie on its circumference.
If GIVEN_LABEL measures GIVEN^\circ,
what does ASKED_LABEL measure?
^\circ
CENTRAL_LABEL and ANGLE_B_LABEL are supplementary.
\begin{eqnarray}
CENTRAL_LABEL + ANGLE_B_LABEL &=& 180^\circ \\
ANGLE_B_LABEL &=& 180^\circ - CENTRAL_LABEL \\
ANGLE_B_LABEL &=& 180^\circ - \blue{CENTRAL^\circ} \\
ANGLE_B_LABEL &=& \green{180 - CENTRAL^\circ}
\end{eqnarray}
The angles in a triangle sum to 180^\circ.
ANGLE_B_LABEL + ANGLE_C_LABEL + ANGLE_D_LABEL = 180^\circ
The pink sides of triangle CBD are radii, so they must be equal.
This means triangle CBD is isosceles and that the base angles,
ANGLE_C_LABEL and ANGLE_D_LABEL angles, are equal.
\begin{eqnarray}
ANGLE_B_LABEL + 2 \cdot ANGLE_D_LABEL &=& 180^\circ \\
2 \cdot ANGLE_D_LABEL &=& 180^\circ - \green{180 - CENTRAL^\circ} \\
2 \cdot ANGLE_D_LABEL &=& CENTRAL^\circ
\end{eqnarray}
ANGLE_D_LABEL = \orange{CENTRAL / 2^\circ},
(half CENTRAL_LABEL).
^\circ
The pink sides of triangle CBD are radii, so they must be equal.
This means triangle CBD is isosceles and that the base angles,
ANGLE_C_LABEL and ANGLE_D_LABEL angles, are equal.
The angles in a triangle sum to 180^\circ.
ANGLE_B_LABEL + ANGLE_C_LABEL + ANGLE_D_LABEL = 180^\circ
\begin{eqnarray}
ANGLE_B_LABEL + 2 \cdot ANGLE_D_LABEL &=& 180^\circ \\
ANGLE_B_LABEL &=& 180^\circ - 2 \cdot \orange{GIVEN^\circ} \\
ANGLE_B_LABEL &=& \green{180 - CENTRAL^\circ}
\end{eqnarray}
CENTRAL_LABEL and ANGLE_B_LABEL are supplementary.
\begin{eqnarray}
CENTRAL_LABEL + ANGLE_B_LABEL &=& 180^\circ \\
CENTRAL_LABEL &=& 180^\circ - ANGLE_B_LABEL \\
CENTRAL_LABEL &=& 180^\circ - \green{180 - CENTRAL^\circ}
\end{eqnarray}
CENTRAL_LABEL = \blue{CENTRAL^\circ},
(twice GIVEN_LABEL).
First divide the angles into two.
CENTRAL_LABEL = SUBANGLE_1 + SUBANGLE_2
ANGLE_D_LABEL = SUBANGLE_3 + SUBANGLE_4
SUBANGLE_1 and ANGLE_B_LABEL are supplementary.
So, ANGLE_B_LABEL = 180^\circ - SUBANGLE_1
The angles in a triangle sum to 180^\circ.
ANGLE_B_LABEL + ANGLE_C_LABEL + SUBANGLE_2 = 180^\circ
The pink sides of triangle CBD are radii, so they must be equal.
This means triangle CBD is isosceles and that the base angles,
ANGLE_C_LABEL and SUBANGLE_3 angles, are equal.
\begin{eqnarray}
ANGLE_B_LABEL + 2 \cdot SUBANGLE_3 &=& 180^\circ \\
2 \cdot SUBANGLE_3 &=& 180^\circ - ANGLE_B_LABEL \\
2 \cdot SUBANGLE_3 &=& 180^\circ - (180^\circ - SUBANGLE_1) \\
2 \cdot SUBANGLE_3 &=& SUBANGLE_1 \\
SUBANGLE_3 &=& \frac{1}{2}SUBANGLE_1
\end{eqnarray}
By the same logic, SUBANGLE_4 = \frac{1}{2}SUBANGLE_2.
So, ASKED_LABEL = \frac{1}{2}GIVEN_LABEL.
ASKED_LABEL = \frac{1}{2} \cdot \blue{CENTRAL}^\circ = \orange{CENTRAL / 2^\circ}
^\circ
First divide the angles into two.
CENTRAL_LABEL = SUBANGLE_1 + SUBANGLE_2
ANGLE_D_LABEL = SUBANGLE_3 + SUBANGLE_4
The pink sides of triangle CBD are radii, so they must be equal.
This means triangle CBD is isosceles and that the base angles,
ANGLE_C_LABEL and SUBANGLE_3 angles, are equal.
The angles in a triangle sum to 180^\circ.
ANGLE_B_LABEL + ANGLE_C_LABEL + SUBANGLE_3 = 180^\circ
\begin{eqnarray}
ANGLE_B_LABEL + 2 \cdot SUBANGLE_3 &=& 180^\circ \\
ANGLE_B_LABEL &=& 180^\circ - 2 \cdot SUBANGLE_3
\end{eqnarray}
SUBANGLE_1 and ANGLE_B_LABEL are supplementary.
\begin{eqnarray}
SUBANGLE_1 + ANGLE_B_LABEL &=& 180^\circ \\
SUBANGLE_1 &=& 180^\circ - ANGLE_B_LABEL \\
SUBANGLE_1 &=& 180^\circ - (180^\circ - 2 \cdot SUBANGLE_3) \\
SUBANGLE_1 &=& 2 \cdot SUBANGLE_3
\end{eqnarray}
By the same logic, SUBANGLE_2 = 2 \cdot SUBANGLE_4.
So, ASKED_LABEL = 2 \cdot GIVEN_LABEL.
ASKED_LABEL = 2 \cdot \orange{CENTRAL / 2}^\circ = \blue{CENTRAL^\circ}
First add a diameter.
The triangle \pink{ABD} is isosceles as
\pink{AB} and \pink{BD} are radii, so must be equal.
Therefore, \pink{\angle{BAD}} and \orange{\angle{ADB}} are equal.
Since, the angles in a triangle sum to 180^\circ
\begin{eqnarray}
\green{\angle{ABD}} + \pink{\angle{BAD}} + \orange{\angle{ADB}} &=& 180^\circ \\
\green{\angle{ABD}} + 2 \cdot \orange{\angle{ADB}} &=& 180^\circ \\
\green{\angle{ABD}} &=& 180^\circ - 2 \cdot \orange{\angle{ADB}}
\end{eqnarray}
\green{\angle{ABD}} and \blue{\angle{ABE}} are supplementary.
\begin{eqnarray}
\green{\angle{ABD}} + \blue{\angle{ABE}} &=& 180^\circ \\
\blue{\angle{ABE}} &=& 180^\circ - \green{\angle{ABD}} \\
\blue{\angle{ABE}} &=& 180^\circ - (180^\circ - 2 \cdot \orange{\angle{ADB}})
\end{eqnarray}
So, \blue{\angle{ABE}} = 2 \cdot \orange{\angle{ADB}}.
We can also created an isosceles triangle, \pink{BCD}.
Using the same argument we can find that, \blue{\angle{CBE}} = 2 \cdot \orange{\angle{CDB}}.
\blue{\angle{ABC}} = \blue{\angle{ABE}} - \blue{\angle{CBE}}
\orange{\angle{ADC}} = \orange{\angle{ADE}} - \orange{\angle{CDE}}
\blue{\angle{ABC}} = \blue{\angle{CBE}} - \blue{\angle{ABE}}
\orange{\angle{ADC}} = \orange{\angle{CDE}} - \orange{\angle{ADE}}
Since \blue{\angle{ABE}} = 2 \cdot \orange{\angle{ADE}} and
\blue{\angle{CBE}} = 2 \cdot \orange{\angle{CDE}},
\blue{\angle{ABC}} = 2 \cdot \orange{\angle{ADC}}
So, \orange{\angle{ADC}} = \frac{1}{2} \cdot \blue{GIVEN^\circ} = GIVEN / 2^\circ.
So, \blue{\angle{ABC}} = 2 \cdot \orange{GIVEN^\circ} = GIVEN * 2^\circ.
^\circ