arc BD is 25°
Explanation:We would apply the secant-secant theorem:
[tex]\begin{gathered} \angle A\text{ =}\frac{large\text{ arc - small arc}}{2} \\ \angle A\text{ =}\frac{arc\text{ CE - arc BD}}{2} \end{gathered}[/tex]angle A = ∠A =25°
arc BD =?
arc CE = 100°
[tex]\begin{gathered} 25\text{ = }\frac{100-arc\text{ BD}}{2} \\ \text{cross multiply:} \\ 2(25)\text{ = 100 - arc BD} \end{gathered}[/tex][tex]\begin{gathered} 50\text{ = 100 - arc BD} \\ \text{subtract 100 from both sides:} \\ 50\text{ - 100 = 100 - 100 - arc BD} \\ -50\text{ = - arc BD} \end{gathered}[/tex][tex]\begin{gathered} \text{DIvide both sides by -1:} \\ \frac{-50}{-1\text{ }}=\frac{-arc\text{ BD}}{-1} \\ \text{arc BD = 50}\degree \end{gathered}[/tex]