Question

in the figure, \\(\overrightarrow{ba}\\) and \\(\overrightarrow{bc}\\) are opposite rays. \\(\overrightarrow{bh}\\) bisects \\(\angle ebc\\) and \\(\overrightarrow{be}\\) bisects \\(\angle abf\\). if \\(m\angle ebc = (3r + 10)\degree\\) and \\(m\angle abe = (2r - 20)\degree\\), find \\(m\angle ebf\\).

Explanation:

Step1: Identify angle relationship

Since \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\) are opposite rays, \(\angle ABC = 180^\circ\). So \(\angle EBC+\angle ABE = 180^\circ\). Substitute the given expressions: \((3r + 10)+(2r - 20)=180\).

Step2: Solve for \(r\)

Simplify the equation: \(3r+10 + 2r-20 = 180\) → \(5r - 10 = 180\) → \(5r=190\) → \(r = 38\).

Step3: Find \(m\angle EBC\) and \(m\angle ABE\)

Calculate \(m\angle EBC=(3\times38 + 10)^\circ=(114 + 10)^\circ = 124^\circ\).
Calculate \(m\angle ABE=(2\times38 - 20)^\circ=(76 - 20)^\circ = 56^\circ\).

Step4: Use angle bisectors

\(\overrightarrow{BE}\) bisects \(\angle ABF\), so \(\angle ABE=\angle EBF\)? Wait, no: Wait, \(\overrightarrow{BH}\) bisects \(\angle EBC\), and \(\overrightarrow{BE}\) bisects \(\angle ABF\). Wait, actually, since \(\angle ABC = 180^\circ\), and we need \(\angle EBF\). Wait, let's re - examine: \(\overrightarrow{BE}\) bisects \(\angle ABF\) (so \(\angle ABE=\angle EBF\)) and \(\overrightarrow{BH}\) bisects \(\angle EBC\). But we need \(\angle EBF\). Wait, from \(\angle ABE = 56^\circ\), and \(\overrightarrow{BE}\) bisects \(\angle ABF\), so \(\angle EBF=\angle ABE\)? Wait, no, wait the problem: Wait, \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\) are opposite rays, so \(\angle ABC = 180^\circ\). \(\angle EBC+(2r - 20)=180\), we found \(r = 38\), \(\angle ABE = 56^\circ\). Since \(\overrightarrow{BE}\) bisects \(\angle ABF\), then \(\angle EBF=\angle ABE\)? Wait, no, wait: Wait, \(\angle ABF\) is split by \(BE\) into two equal angles, so \(\angle ABE=\angle EBF\). Wait, but also, \(\overrightarrow{BH}\) bisects \(\angle EBC\), but we need \(\angle EBF\). Wait, actually, since \(\angle ABE = 56^\circ\), and \(BE\) bisects \(\angle ABF\), then \(\angle EBF=\angle ABE = 56^\circ\)? Wait, no, that can't be. Wait, no, wait: Wait, \(\angle ABC = 180^\circ\), \(\angle EBC = 124^\circ\), \(\angle ABE = 56^\circ\). \(\overrightarrow{BE}\) bisects \(\angle ABF\), so \(\angle ABF = 2\angle ABE=112^\circ\). \(\overrightarrow{BH}\) bisects \(\angle EBC\), so \(\angle EBH=\frac{1}{2}\angle EBC = 62^\circ\). Wait, maybe I made a mistake. Wait, the problem says "find \(m\angle EBF\)". Wait, let's re - read the problem: " \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\) are opposite rays. \(\overrightarrow{BH}\) bisects \(\angle EBC\) and \(\overrightarrow{BE}\) bisects \(\angle ABF\). Find \(m\angle EBF\)".

Wait, since \(\angle ABC = 180^\circ\), \(\angle EBC+\angle ABE = 180^\circ\). We found \(r = 38\), so \(\angle ABE=(2\times38 - 20)=56^\circ\). Since \(\overrightarrow{BE}\) bisects \(\angle ABF\), then \(\angle EBF=\angle ABE = 56^\circ\)? Wait, no, that would mean \(\angle ABF = 112^\circ\), and \(\angle EBC = 124^\circ\), and \(\angle ABC=180^\circ\), which is consistent because \(56 + 124=180\). Wait, but then \(\angle EBF=\angle ABE = 56^\circ\)? Wait, no, maybe I messed up the bisector. Wait, \(\overrightarrow{BE}\) bisects \(\angle ABF\), so \(\angle ABE=\angle EBF\), so \(\angle EBF = 56^\circ\)? Wait, but let's check again.

Wait, step - by - step:

1. \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\) are opposite rays, so \(\angle ABC = 180^\circ\). So \(\angle EBC+\angle ABE=180^\circ\).

2. Given \(m\angle EBC=(3r + 10)^\circ\) and \(m\angle ABE=(2r - 20)^\circ\). So:

\((3r + 10)+(2r - 20)=180\)

\(5r-10 = 180\)

\(5r=190\)

\(r = 38\)

3. Then \(m\angle ABE=(2\times38 - 20)=56^\circ\)

4. Since \(\overrightarrow{BE}\) bisects \(\angle ABF\), then \(\angle EBF=\angle ABE\) (by definition of angle bisector: a bisector divides an angle into two equal parts). So \(m\angle EBF = 56^\circ\)? Wait, but that seems too small. Wait, no, wait \(\angle ABF\) is part of \(\angle ABC\)? Wait, no, \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\) are opposite rays, so \(\angle ABC = 180^\circ\), \(\angle ABF\) is between \(\overrightarrow{BA}\) and \(\overrightarrow{BF}\), and \(\overrightarrow{BE}\) bisects \(\angle ABF\), so \(\angle ABE=\angle EBF\). And \(\angle EBC\) is between \(\overrightarrow{BE}\) and \(\overrightarrow{BC}\), with \(\angle EBC = 124^\circ\), \(\angle ABE = 56^\circ\), and \(56 + 124 = 180\), which is correct. So \(\angle EBF=\angle ABE = 56^\circ\)? Wait, but maybe I misread the bisector. Wait, the problem says "\(\overrightarrow{BH}\) bisects \(\angle EBC\) and \(\overrightarrow{BE}\) bisects \(\angle ABF\)". So \(\overrightarrow{BE}\) bisects \(\angle ABF\), so \(\angle ABE=\angle EBF\), so \(m\angle EBF = 56^\circ\). Wait, but let's check again.

Wait, another way: \(\angle ABC = 180^\circ\), \(\angle EBC = 124^\circ\), \(\angle ABE = 56^\circ\). \(\overrightarrow{BE}\) bisects \(\angle ABF\), so \(\angle ABF = 2\angle ABE = 112^\circ\), so \(\angle EBF=\angle ABE = 56^\circ\). Yes, that makes sense.

Answer:

\(56^\circ\)