For problem 42, we are given:

• $m \angle EBC = 3r + 10$
• $m \angle ABE = 2r - 20$

Since $BH$ bisects $\angle EBC$, we know:

$m \angle EBF = m \angle EBC - m \angle ABE$

Setting up the equation and solving, we find that:

• $r = -30$

Substituting $r = -30$ into the equation for $m \angle EBF$, we find that:

$m \angle EBF = 0^\circ$

Would you like to see more details about the solution process or ask any further questions?

Here are some related questions you might explore:

1. How do angle bisectors work in geometry?
2. Why is $m \angle EBF$ equal to 0 degrees in this case?
3. Can you explain the significance of opposite rays in this problem?
4. What happens if $r$ is positive? Is the setup always consistent?
5. How does the equation change if the angle bisector condition is modified?

Tip: Always double-check the logic of geometric constraints when solving equations involving bisected angles.