Question

in the figure, $overrightarrow{ba}$ and $overrightarrow{bc}$ are opposite rays. $overrightarrow{bh}$ bisects $angle ebc$ and $overrightarrow{be}$ bisects $angle abf$. if $mangle abf = (7b - 24)^circ$ and $mangle abe = 2b^circ$, find $mangle ebf$.

Explanation:

Step1: Use angle bisector property

Since \( \overrightarrow{BE} \) bisects \( \angle ABF \), we know that \( m\angle ABE = m\angle EBF \), and also \( m\angle ABF = m\angle ABE + m\angle EBF = 2m\angle ABE \) (because \( m\angle ABE = m\angle EBF \)).
Given \( m\angle ABF=(7b - 24)^{\circ} \) and \( m\angle ABE = 2b^{\circ} \), we can set up the equation:
\( 7b-24 = 2\times(2b) \)

Step2: Solve for \( b \)

Simplify the right - hand side of the equation: \( 7b-24 = 4b \)
Subtract \( 4b \) from both sides: \( 7b-4b-24=4b - 4b \), which gives \( 3b-24 = 0 \)
Add 24 to both sides: \( 3b-24 + 24=0 + 24 \), so \( 3b=24 \)
Divide both sides by 3: \( b=\frac{24}{3}=8 \)

Step3: Find \( m\angle ABE \)

Substitute \( b = 8 \) into the expression for \( m\angle ABE \): \( m\angle ABE=2b^{\circ}=2\times8^{\circ}=16^{\circ} \)

Step4: Find \( m\angle EBF \)

Since \( \overrightarrow{BE} \) bisects \( \angle ABF \), \( m\angle EBF=m\angle ABE \). So \( m\angle EBF = 16^{\circ} \) (We can also verify using \( m\angle ABF \): \( m\angle ABF=7b - 24=7\times8-24=56 - 24 = 32^{\circ} \), and \( m\angle EBF=\frac{1}{2}m\angle ABF=\frac{32^{\circ}}{2}=16^{\circ} \))

Answer:

\( 16 \)