Question

find ( moverarc{cd} ).
( moverarc{cd} = square^circ )

Explanation:

Step1: Recall the property of a cyclic quadrilateral

The sum of opposite angles in a cyclic quadrilateral is \(180^{\circ}\). Also, the sum of the measures of the arcs of a circle is \(360^{\circ}\). Let \(m\overset{\frown}{CD}=x\).

Step2: Use the angle - arc relationship

We know that the sum of the measures of the interior angles of a cyclic quadrilateral \(A + B+C + D=360^{\circ}\), and also \(m\overset{\frown}{A}+m\overset{\frown}{B}+m\overset{\frown}{C}+m\overset{\frown}{D}=360^{\circ}\). But another way is:
The sum of the measures of the arcs of a circle is \(360^{\circ}\). We know that \(m\overset{\frown}{AB}+m\overset{\frown}{BC}+m\overset{\frown}{CD}+m\overset{\frown}{DA}=360^{\circ}\).
We also know that for a cyclic quadrilateral \(A + C=180^{\circ}\) and \(B + D = 180^{\circ}\). But using the arc - angle relationship:
The measure of an inscribed angle \(I=\frac{1}{2}\) (measure of its intercepted arc). But another formula: If \(A\), \(B\), \(C\), \(D\) are vertices of a cyclic quadrilateral, then \(m\overset{\frown}{AB}+m\overset{\frown}{CD}=2(A + C)\) and \(m\overset{\frown}{BC}+m\overset{\frown}{DA}=2(B + D)\). Since \(A + C=180^{\circ}\) (property of cyclic quadrilateral), \(m\overset{\frown}{AB}+m\overset{\frown}{CD}=360^{\circ}\).
We are given \(m\overset{\frown}{AB}\) (indirectly, we can calculate using the fact that the sum of arcs: Let's use the formula \(m\overset{\frown}{AB}+m\overset{\frown}{BC}+m\overset{\frown}{CD}+m\overset{\frown}{DA}=360^{\circ}\).
We know that \(m\overset{\frown}{BC} = 112^{\circ}\).
We use the property that \(A=\frac{1}{2}(m\overset{\frown}{BCD})\) (where \(m\overset{\frown}{BCD}=m\overset{\frown}{BC}+m\overset{\frown}{CD}\)) and \(B=\frac{1}{2}(m\overset{\frown}{CDA})\) (not the easiest way).
The best formula is: The sum of the measures of the arcs of a circle is \(360^{\circ}\).
We know that \(m\overset{\frown}{AB}\) (if we use the property of cyclic quadrilateral \(A + C=180^{\circ}\), \(B + D=180^{\circ}\). But using the arc sum:
\(m\overset{\frown}{CD}=360-(112 + 131+103)\) is wrong.
The correct formula: The sum of the measures of the opposite arcs of a cyclic quadrilateral is \(360^{\circ}\). Wait, no. The sum of all arcs in a circle is \(360^{\circ}\).
We use the property that \(m\angle A=\frac{1}{2}(m\overset{\frown}{BCD})\), \(m\angle B=\frac{1}{2}(m\overset{\frown}{CDA})\), \(m\angle C=\frac{1}{2}(m\overset{\frown}{DAB})\), \(m\angle D=\frac{1}{2}(m\overset{\frown}{ABC})\).
Another property: The sum of the measures of the interior angles of a cyclic quadrilateral \(A + B + C+D = 360^{\circ}\), and \(m\overset{\frown}{AB}+m\overset{\frown}{BC}+m\overset{\frown}{CD}+m\overset{\frown}{DA}=360^{\circ}\)
We know that \(m\overset{\frown}{AB}=2(180 - 131)= 98^{\circ}\) (because the inscribed angle \(D=\frac{1}{2}m\overset{\frown}{AB}\), and \(A + C=180^{\circ}\), \(B + D=180^{\circ}\), so \(D = 180 - 103=77^{\circ}\), then \(m\overset{\frown}{AB}=2D = 154^{\circ}\) (wrong approach).
The correct formula:
We know that \(m\overset{\frown}{AB}+m\overset{\frown}{BC}+m\overset{\frown}{CD}+m\overset{\frown}{DA}=360^{\circ}\)
We use the property that \(m\angle A=\frac{1}{2}(m\overset{\frown}{BCD})\), \(m\angle B=\frac{1}{2}(m\overset{\frown}{CDA})\)
Since \(A = 131^{\circ}\), \(m\overset{\frown}{BCD}=2\times131=262^{\circ}\) (where \(m\overset{\frown}{BCD}=m\overset{\frown}{BC}+m\overset{\frown}{CD}\))
Given \(m\overset{\frown}{BC} = 112^{\circ}\)
We solve for \(m\overset{\frown}{CD}\):
\(m\overset{\frown}{CD}=2\times131 - 112\)

Step3: Calculate \(m\overset{\frown}{CD}\)

\[ \begin{align*} m\overset{\frown}{CD}&=262-112\\ &=150 \end{align*} \]

Answer:

\(150\)