Question

triangle bcd is isosceles and ( overline{bc} cong overline{bd} ). what is the measure of ( overparen{bc} )?

Explanation:

Step1: Find the measure of ∠CBD

The measure of an inscribed angle is half the measure of its intercepted arc. The arc \(CD\) is \(100^{\circ}\), so \(\angle CBD=\frac{1}{2}\times100^{\circ} = 50^{\circ}\)

Step2: Use the property of isosceles triangle

Since \(\triangle BCD\) is isosceles with \(BC = BD\), \(\angle BCD=\angle BDC\). The sum of angles in a triangle is \(180^{\circ}\). Let \(\angle BCD=\angle BDC = x\). Then \(x + x+50^{\circ}=180^{\circ}\), \(2x=130^{\circ}\), \(x = 65^{\circ}\)

Step3: Find the measure of arc \(BC\)

The measure of an inscribed angle \(\angle BDC\) is half the measure of arc \(BC\). Let the measure of arc \(BC\) be \(y\). Since \(\angle BDC = 65^{\circ}\), then \(y = 2\times65^{\circ}=130^{\circ}\)

Answer:

\(130^{\circ}\)