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The given diagram shows that \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\) are opposite rays, and line segments \(BH\) and \(BE\) bisect \(\angle EBC\) and \(\angle ABF\), respectively.

We are given the following information:
1. \( m\angle EBC = (31a - 2)^\circ \)
2. \( m\angle EBH = (4a + 45)^\circ \)

Our goal is to find \( m\angle HBC \).

### Step-by-Step Solution:

#### 1. Use the fact that \(BH\) bisects \(\angle EBC\):
Since \(BH\) bisects \(\angle EBC\), it means that:
\[
m\angle EBH = m\angle HBC
\]
Thus, we have:
\[
m\angle HBC = (4a + 45)^\circ
\]

#### 2. Use the full expression for \( m\angle EBC \):
We know:
\[
m\angle EBC = m\angle EBH + m\angle HBC
\]
Substituting the known values, we get:
\[
(31a - 2)^\circ = (4a + 45)^\circ + (4a + 45)^\circ
\]
Simplify the right-hand side:
\[
(31a - 2)^\circ = 2(4a + 45)^\circ
\]
\[
(31a - 2)^\circ = (8a + 90)^\circ
\]

#### 3. Solve for \(a\):
Equating the two expressions:
\[
31a - 2 = 8a + 90
\]
Now, solve for \(a\):
\[
31a - 8a = 90 + 2
\]
\[
23a = 92
\]
\[
a = 4
\]

#### 4. Find \(m\angle HBC\):
Substitute \(a = 4\) into the expression for \(m\angle HBC\):
\[
m\angle HBC = 4a + 45 = 4(4) + 45 = 16 + 45 = 61^\circ
\]

### Final Answer:
\[
m\angle HBC = 61^\circ
\]

Would you like further clarifications or details? Here are five related questions to consider:
1. How would the solution change if \(BH\) did not bisect \(\angle EBC\)?
2. Can you find the value of \(m\angle ABF\) from the given information?
3. How is the concept of angle bisectors used in geometric constructions?
4. What other angle relationships can be derived from this figure?
5. How can you verify the consistency of your solution with geometric properties?

**Tip:** Always check if any bisected angles or linear pairs can help simplify the problem through symmetry.

In the figure, BA and BC are opposite rays. BH bisects ∠EBC and BE bisects ∠ABF. If m∠EBC = (31a - 2)° and m∠EBH = (4a + 45)°, find m∠HBC.

Learn how to solve a geometry problem involving opposite rays and bisected angles. This step-by-step solution finds the measure of angle HBC when m∠EBC = (31a - 2)° and m∠EBH = (4a + 45)°. Suitable for Grades 9-12 students.

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