Question

triangle bcd is isosceles and $overline{bc} cong overline{bd}$. what is the measure of $overset{\frown}{bc}$? $\bigcirc 100^circ$ $\bigcirc 120^circ$ $\bigcirc 130^circ$ $\bigcirc 160^circ$

Explanation:

Step1: Find base angles of $\triangle BCD$

Since $\triangle BCD$ is isosceles with $\overline{BC} \cong \overline{BD}$, $\angle BCD = \angle BDC = 100^\circ$ is impossible (sum exceeds $180^\circ$). Correct: $\angle CBD$ is vertex angle, so $\angle BCD = \angle BDC$.
$\angle CBD = 180^\circ - 2\times100^\circ$ is invalid, so $\angle BDC = 100^\circ$ is wrong. Actually, $\angle BCD = \angle CBD$:
$\angle BDC = 180^\circ - 2\times\angle BCD$. Wait, given $\angle BDC = 100^\circ$? No, the $100^\circ$ is $\angle BCD$. Correct:
$\triangle BCD$: $\overline{BC} \cong \overline{BD}$, so base angles $\angle BCD = \angle BDC$. Given $\angle BCD = 100^\circ$ is wrong, so $\angle BDC = 100^\circ$, then $\angle CBD = 180^\circ - 2\times100^\circ$ invalid. **Correct**: The inscribed angle $\angle BDC = 100^\circ$ is not, the inscribed angle subtended by arc $BC$ is $\angle BDC$. Wait, inscribed angle theorem: inscribed angle = $\frac{1}{2}$ measure of its subtended arc.

Step2: Calculate $\angle CBD$

$\triangle BCD$: $\overline{BC} \cong \overline{BD}$, so $\angle BCD = \angle BDC$. Let $\angle BCD = \angle BDC = x$. Given the external? No, the $100^\circ$ is $\angle BCD$. Wait, sum of angles in triangle:
$\angle CBD = 180^\circ - 2\times100^\circ$ is invalid, so the $100^\circ$ is $\angle BDC$, so $\angle BCD = \angle CBD = x$:
$x + x + 100^\circ = 180^\circ$
$2x = 80^\circ$
$x = 40^\circ$
So $\angle BCD = 40^\circ$

Step3: Relate inscribed angle to arc

Inscribed angle $\angle BDC$ subtends arc $BC$. Wait, no: inscribed angle $\angle BDC$ subtends arc $BC$, so $m\angle BDC = \frac{1}{2}m\widehat{BC}$. Wait, no: $\angle BDC$ is an inscribed angle subtended by arc $BC$, so $m\widehat{BC} = 2\times m\angle BDC$. But $\angle BDC = 100^\circ$? No, $\angle BDC = 80^\circ$? Wait, no, $\angle CBD = 40^\circ$, which is an inscribed angle subtended by arc $CD$. Wait, correct: $\angle BDC$ is inscribed angle subtended by arc $BC$, so $m\widehat{BC} = 2\times m\angle BDC$. But $\angle BDC = 100^\circ$ is wrong, because triangle angles can't exceed $180^\circ$. **Correct approach**:
The $100^\circ$ is $\angle BCD$, $\triangle BCD$ is isosceles with $\overline{BC} \cong \overline{BD}$, so $\angle BCD = \angle BDC = 100^\circ$ is impossible, so the $100^\circ$ is $\angle BDC$, so $\angle BCD = \angle CBD = \frac{180^\circ - 100^\circ}{2} = 40^\circ$.
Now, $\angle BCD$ is an inscribed angle subtended by arc $BD$, so $m\widehat{BD} = 2\times40^\circ = 80^\circ$. Since $\overline{BC} \cong \overline{BD}$, arcs $\widehat{BC} \cong \widehat{BD}$, so $m\widehat{BC} = 80^\circ$? No, that's not an option. Wait, no: $\angle BDC$ is an inscribed angle subtended by arc $BC$, so $m\widehat{BC} = 2\times m\angle BDC$. But $\angle BDC$ is not $100^\circ$, the $100^\circ$ is the reflex? No, the total circle is $360^\circ$. Wait, the inscribed angle $\angle BCD = 100^\circ$ is a reflex? No, inscribed angle can't be more than $180^\circ$, but the subtended arc is the major arc. So if $\angle BCD = 100^\circ$, it subtends the major arc $BD$, so major arc $BD = 2\times100^\circ = 200^\circ$, so minor arc $BD = 360^\circ - 200^\circ = 160^\circ$. Since $\overline{BC} \cong \overline{BD}$, minor arc $BC =$ minor arc $BD = 160^\circ$? No, that's an option. Wait, no: $\triangle BCD$ is isosceles with $\overline{BC} \cong \overline{BD}$, so minor arcs $\widehat{BC} \cong \widehat{BD}$. The inscribed angle $\angle BCD$ subtends arc $BD$, so if $\angle BCD = 100^\circ$, that is the inscribed angle for major arc $BD$, so major arc $BD = 2\times100^\circ = 200^\circ$, so minor arc $BD = 360^\circ - 200^\circ = 160^\circ$, so minor arc $BC = 160^\circ$.

Step4: Confirm with triangle angles

If minor arc $BC = 160^\circ$, then inscribed angle $\angle BDC = \frac{1}{2}\times160^\circ = 80^\circ$. Then $\triangle BCD$: $\angle BCD = \angle BDC = 80^\circ$, $\angle CBD = 180^\circ - 80^\circ - 80^\circ = 20^\circ$, which fits. The $100^\circ$ is the reflex of $\angle BCD$? No, the diagram shows $100^\circ$ at $\angle BCD$, which is the external? No, the $100^\circ$ is the supplementary angle of $\angle BCD$, so $\angle BCD = 80^\circ$, which matches.

Answer:

D. $160^\circ$