Question

in \\( \odot o, m \widehat { b d } = 130 ^ { \circ } \\) and \\( \overline { a b } \cong \overline { c d } \\). also, the center of the circle, point \\( o \\), is the intersection of \\( \overline { c b } \\) and \\( \overline { a d } \\).
what is \\( m \angle 4 \\)?
\\( m \angle 4 = \square ^ { \circ } \\)

Explanation:

Step1: Use the property of congruent chords

Since \(\overline{AB}\cong\overline{CD}\), then \(\angle2=\angle1\) and \(\angle3 = \angle4\).

Step2: Calculate the sum of \(\angle3\) and \(\angle4\)

The measure of arc \(BD\) is \(m\widehat{BD}=130^{\circ}\). The central - angle theorem states that the measure of an arc is equal to the measure of its central angle. So \(m\angle BOD = 130^{\circ}\). Also, \(\angle3+\angle4=180^{\circ}-m\angle BOD\). Substitute \(m\angle BOD = 130^{\circ}\) into the formula: \(\angle3+\angle4=180 - 130=50^{\circ}\).

Step3: Solve for \(\angle4\)

Because \(\angle3=\angle4\) (from Step 1), let \(x = \angle4\). Then \(x + x=50^{\circ}\), \(2x = 50^{\circ}\), and \(x=\frac{50^{\circ}}{2}=25^{\circ}\).

Answer:

\(25\)