Question

triangle bcd is isosceles and bc ≅ bd. what is the measure of arc bc? 100° 120° 130° 160°

Explanation:

Step1: Recall the inscribed - angle theorem

The measure of an inscribed angle is half the measure of its intercepted arc. The inscribed angle $\angle BCD$ intercepts arc $\overset{\frown}{BD}$, and $\angle BCD = 100^{\circ}$, so the measure of arc $\overset{\frown}{BD}=2\times\angle BCD = 200^{\circ}$.

Step2: Use the property of the sum of arcs in a circle

The sum of the measures of the arcs of a circle is $360^{\circ}$. Let the measure of arc $\overset{\frown}{BC}$ be $x$. Since $\overline{BC}\cong\overline{BD}$, the measure of arc $\overset{\frown}{BC}$ is equal to the measure of arc $\overset{\frown}{BD}$.
We know that $x + x+ \text{(the remaining arc)}=360^{\circ}$. The remaining arc corresponding to $\angle BCD$ has measure $200^{\circ}$. So $2x+200^{\circ}=360^{\circ}$.

Step3: Solve for $x$

Subtract $200^{\circ}$ from both sides of the equation $2x + 200^{\circ}=360^{\circ}$:
$2x=360^{\circ}- 200^{\circ}=160^{\circ}$.
Divide both sides by 2: $x = 160^{\circ}\div2 = 160^{\circ}$.

Answer:

$160^{\circ}$