In the figure, BA−→BA→ and BC−→−BC→ are opposite rays. BH−→−BH→ bisects ∠EBC∠EBC and BE−→BE→ bisects ∠ABF∠ABF.

 
If m∠EBC=(3r+10)°m∠EBC=3r+10° and m∠ABE=(2r−20)°m∠ABE=2r-20°, find m∠EBFm∠EBF.
 

We are given the following information:

- \( \overrightarrow{BA} \) and \( \overrightarrow{BC} \) are opposite rays, meaning \( \angle ABC = 180^\circ \).
- \( \overrightarrow{BH} \) bisects \( \angle EBC \), meaning \( \angle EBC = 2 \times \angle HBC \).
- \( \overrightarrow{BE} \) bisects \( \angle ABF \), meaning \( \angle ABF = 2 \times \angle ABE \).

### Information about the angles:
- \( m\angle EBC = (3r + 10)^\circ \)
- \( m\angle ABE = (2r - 20)^\circ \)

### Step 1: Express \( m\angle ABC \)
Since \( \overrightarrow{

In the figure, BA−→ and BC−→ are opposite rays. BH−→ bisects ∠EBC and BE−→ bisects ∠ABF. If m∠EBC=(3r+10)° and m∠ABE=(2r−20)°, find m∠EBF.

Learn how to find m∠EBF given m∠EBC = (3r + 10)° and m∠ABE = (2r − 20)°. This problem involves geometric concepts like angle bisectors, linear pair of angles, and the use of the linear pair theorem. Ideal for students in Grades 9-10.

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