Angle Bisector A32742

1. **State the problem:** We are given that $\overrightarrow{BA}$ and $\overrightarrow{BC}$ are opposite rays, $\overrightarrow{BH}$ bisects $\angle EBC$, and $\overrightarrow{BE}$ bisects $\angle ABF$. We know $m\angle EBC = (31a - 2)^\circ$ and $m\angle EBH = (4a + 45)^\circ$. We need to find $m\angle HBC$. 2. **Understand the angle bisector:** Since $\overrightarrow{BH}$ bisects $\angle EBC$, it divides $\angle EBC$ into two equal angles: $$m\angle EBH = m\angle HBC$$ 3. **Set up the equation:** Because $\angle EBC$ is split into two equal parts by $\overrightarrow{BH}$, we have: $$m\angle EBC = m\angle EBH + m\angle HBC = 2 \times m\angle EBH$$ 4. **Substitute the given expressions:** $$31a - 2 = 2(4a + 45)$$ 5. **Simplify and solve for $a$:** $$31a - 2 = 8a + 90$$ $$31a - 8a = 90 + 2$$ $$23a = 92$$ $$a = \frac{92}{23} = 4$$ 6. **Find $m\angle EBH$ using $a=4$:** $$m\angle EBH = 4a + 45 = 4(4) + 45 = 16 + 45 = 61^\circ$$ 7. **Find $m\angle HBC$:** Since $\overrightarrow{BH}$ bisects $\angle EBC$, $$m\angle HBC = m\angle EBH = 61^\circ$$ **Final answer:** $$\boxed{61^\circ}$$