Question

in the figure, \\( \overrightarrow { b a } \\) and \\( \overrightarrow { b c } \\) are opposite rays. \\( \overrightarrow { b h } \\) bisects \\( \angle e b c \\) and \\( \overrightarrow { b e } \\) bisects \\( \angle a b f \\).
if \\( m \angle a b e = ( 2 n + 7 ) ^ { \circ } \\) and \\( m \angle e b f = ( 4 n - 1 ) ^ { \circ } \\), find \\( m \angle a b h \\).

Explanation:

Step1: Use the angle - bisector property

Since \(\overrightarrow{BE}\) bisects \(\angle ABF\), then \(m\angle ABE=m\angle EBF\).
We have the equation \(2n + 7=4n-11\).

Step2: Solve the equation for \(n\)

Subtract \(2n\) from both sides: \(2n+7 - 2n=4n - 11-2n\), which gives \(7 = 2n-11\).
Add \(11\) to both sides: \(7 + 11=2n-11 + 11\), so \(18 = 2n\).
Divide both sides by \(2\): \(n=\frac{18}{2}=9\).

Step3: Find \(m\angle ABF\)

Since \(m\angle ABF=m\angle ABE + m\angle EBF\) and \(m\angle ABE=m\angle EBF\) (because of the angle - bisector), \(m\angle ABF = 2m\angle ABE\).
Substitute \(n = 9\) into \(m\angle ABE=(2n + 7)^{\circ}\). Then \(m\angle ABE=(2\times9 + 7)^{\circ}=(18 + 7)^{\circ}=25^{\circ}\).
So \(m\angle ABF=2\times25^{\circ}=50^{\circ}\).

Step4: Use the fact that \(\angle ABF\) and \(\angle FBC\) are supplementary (since \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\) are opposite rays, \(\angle ABC = 180^{\circ}\))

Let \(m\angle FBC=x\). Then \(m\angle ABF+x = 180^{\circ}\), so \(x=180^{\circ}-m\angle ABF\).
Since \(\overrightarrow{BH}\) bisects \(\angle FBC\), \(m\angle FBH=\frac{1}{2}m\angle FBC\).
First, \(m\angle FBC=180^{\circ}-50^{\circ}=130^{\circ}\). Then \(m\angle FBH=\frac{1}{2}\times130^{\circ}=65^{\circ}\).

Step5: Find \(m\angle ABH\)

\(m\angle ABH=m\angle ABF+m\angle FBH\).
Substitute \(m\angle ABF = 50^{\circ}\) and \(m\angle FBH = 65^{\circ}\). Then \(m\angle ABH=50^{\circ}+65^{\circ}=115^{\circ}\).

Answer:

\(115^{\circ}\)