Question

explore the properties of angles formed by two intersecting chords.

1. the sum of the angle measures was equal to the sum of the arc measures. move points b and c to change the angles and arcs.

make a conjecture. is the sum of the angle measures always equal to the sum of the arc measures?
m∠deb = 105°
m∠aec = 105°
m(widehat{ac}) = 130°
m(widehat{bd}) = 80°
mbd + mac = 210°

Explanation:

Step1: Recall angle - arc relationship

The measure of an angle formed by two intersecting chords in a circle is half the sum of the measures of the intercepted arcs. Let the two intersecting chords be \(AC\) and \(BD\) intersecting at point \(E\). \(\angle DEB\) and \(\angle AEC\) are vertical angles, and \(\angle DEB=\frac{1}{2}(m\overparen{BD}+m\overparen{AC})\), \(\angle AEC=\frac{1}{2}(m\overparen{BD}+m\overparen{AC})\).

Step2: Calculate sum of angles and arcs

Given \(m\angle DEB = 105^{\circ}\), \(m\angle AEC=105^{\circ}\), \(m\overparen{AC} = 130^{\circ}\), \(m\overparen{BD}=80^{\circ}\). The sum of the arc - measures \(m\overparen{BD}+m\overparen{AC}=80^{\circ}+130^{\circ}=210^{\circ}\). The sum of the angle - measures \(m\angle DEB + m\angle AEC=105^{\circ}+105^{\circ}=210^{\circ}\).

Step3: Make a conjecture

In general, for two intersecting chords in a circle, if the angles formed are vertical angles \(\alpha\) and \(\alpha\) (since vertical angles are equal) and the intercepted arcs are \(x\) and \(y\), \(\alpha=\frac{1}{2}(x + y)\), so \(2\alpha=x + y\), which means the sum of the angle measures (the sum of the two vertical angles) is equal to the sum of the arc measures.

Answer:

Yes