Question

in \\( \odot o, m \overgroup{cd}=35^{\circ} \\) and \\( \overline{ca} \cong \overline{bd} \\). also, the center of the circle, point \\( o \\), is the intersection of \\( \overline{cb} \\) and \\( \overline{ad} \\).
what is \\( m \angle 1 \\)?
\\( m \angle 1=\square^{\circ} \\)

Explanation:

Step1: Use the property of congruent chords

Since \(\overline{CA}\cong\overline{BD}\), then \(\overset{\frown}{CA}\cong\overset{\frown}{BD}\). Let \(m\overset{\frown}{CA} = m\overset{\frown}{BD}=x\).

Step2: Use the property of vertical angles and arc - angle relationship

We know that the measure of an inscribed angle is half the measure of its intercepted arc. Also, \(\angle1\) is related to the arcs. The formula for the measure of an angle formed by two chords intersecting at a point (here \(O\)) is \(m\angle1=\frac{1}{2}(m\overset{\frown}{CD}+m\overset{\frown}{AB})\).
Since \(m\overset{\frown}{CA} = m\overset{\frown}{BD}\), we can find that \(m\overset{\frown}{AB}=m\overset{\frown}{CD}\) (by the property of the circle and congruent chords). Given \(m\overset{\frown}{CD} = 35^{\circ}\).

Step3: Calculate \(m\angle1\)

Using the formula \(m\angle1=\frac{1}{2}(m\overset{\frown}{CD}+m\overset{\frown}{AB})\), substituting \(m\overset{\frown}{AB}=m\overset{\frown}{CD} = 35^{\circ}\), we get \(m\angle1=\frac{1}{2}(35^{\circ}+ 35^{\circ})\).

$$m\angle1=\frac{1}{2}\times70^{\circ}=35^{\circ}$$

Answer:

\(35\)