1. **Problem statement:** Given rays $\overrightarrow{OC}$, $\overrightarrow{OD}$, $\overrightarrow{OE}$, $\overrightarrow{OF}$, and $\overrightarrow{OG}$ bisecting angles $\angle AOB$, $\angle AOC$, $\angle AOD$, $\angle AOE$, and $\angle FOC$ respectively, find: a. $m\angle DOE$ if $m\angle BOF = 120$. b. $m\angle EOG$ if $m\angle COG = 35$. 2. **Key concept:** An angle bisector divides an angle into two equal parts. If a ray bisects an angle, the two smaller angles formed are equal. 3. **Step a:** Find $m\angle DOE$ given $m\angle BOF = 120$. - Since $\overrightarrow{OF}$ bisects $\angle AOE$, and $\overrightarrow{OE}$ bisects $\angle AOD$, and $\overrightarrow{OD}$ bisects $\angle AOC$, and $\overrightarrow{OC}$ bisects $\angle AOB$, the angles are successively halved. - $\angle BOF$ is composed of $\angle BOC + \angle COF$. - Because $\overrightarrow{OC}$ bisects $\angle AOB$, $m\angle BOC = \frac{1}{2} m\angle AOB$. - Similarly, $\overrightarrow{OD}$ bisects $\angle AOC$, so $m\angle COD = \frac{1}{2} m\angle AOC$, and so on. - The chain of bisections means each subsequent angle is half the previous. - $m\angle BOF = m\angle BOC + m\angle COF$. - $m\angle COF = m\angle COD + m\angle DOF$, and so forth. - Since each bisector halves the angle, $m\angle BOF = \frac{3}{8} m\angle AOB$ (because after 3 bisections, the angle is $\frac{1}{8}$ of the original, and summing parts gives $\frac{3}{8}$). - Given $m\angle BOF = 120$, then $m\angle AOB = \frac{120 \times 8}{3} = 320$. - $m\angle AOD$ is half of $m\angle AOB$ because $\overrightarrow{OC}$ bisects $\angle AOB$ and $\overrightarrow{OD}$ bisects $\angle AOC$. - So $m\angle AOD = \frac{1}{2} m\angle AOC = \frac{1}{2} \times \frac{1}{2} m\angle AOB = \frac{1}{4} \times 320 = 80$. - $\overrightarrow{OE}$ bisects $\angle AOD$, so $m\angle DOE = \frac{1}{2} m\angle AOD = \frac{1}{2} \times 80 = 40$. 4. **Step b:** Find $m\angle EOG$ given $m\angle COG = 35$. - $\overrightarrow{OG}$ bisects $\angle FOC$, so $m\angle COG = m\angle FOG = 35$. - $\overrightarrow{OF}$ bisects $\angle AOE$, so $m\angle FOE = m\angle FOE = \frac{1}{2} m\angle AOE$. - $\overrightarrow{OE}$ bisects $\angle AOD$, so $m\angle EOD = \frac{1}{2} m\angle AOD$. - $m\angle EOG = m\angle EOD + m\angle DOG$. - Since $\overrightarrow{OG}$ bisects $\angle FOC$, $m\angle DOG = m\angle FOG = 35$. - $m\angle EOD = \frac{1}{2} m\angle AOD$, but without $m\angle AOD$ given, we use the relation from part a or assume the same pattern. - Using the previous result, $m\angle AOD = 80$, so $m\angle EOD = 40$. - Therefore, $m\angle EOG = 40 + 35 = 75$. **Final answers:** $$m\angle DOE = 40$$ $$m\angle EOG = 75$$