1. use the diagram below to complete each part. • ( overline{bf} perp overline{ac} )
a) name the vertex of ( angle 4 ).
b) name the sides of ( angle 1 ).
c) write another name for ( angle 5 ).
d) classify each angle: ( angle fbc ): ( angle ebf ): ( angle abc ):
e) name an angle bisector.
f) if ( mangle ebd = 36^circ ) and ( mangle dbc = 108^circ ), find ( mangle ebc ).
g) if ( mangle ebf = 117^circ ), find ( mangle abe ).

Question

1. use the diagram below to complete each part. • ( overline{bf} perp overline{ac} )
 a) name the vertex of ( angle 4 ).
 b) name the sides of ( angle 1 ).
 c) write another name for ( angle 5 ).
 d) classify each angle: ( angle fbc ): ( angle ebf ): ( angle abc ):
 e) name an angle bisector.
 f) if ( mangle ebd = 36^circ ) and ( mangle dbc = 108^circ ), find ( mangle ebc ).
 g) if ( mangle ebf = 117^circ ), find ( mangle abe ).

Accepted Answer

### Part a)
# Explanation:
## Step1: Recall vertex definition
The vertex of an angle is the common endpoint of its two sides. For $\angle 4$, the sides are $BA$ and $BE$, so the common endpoint (vertex) is $B$.
# Answer:
The vertex of $\angle 4$ is $B$.

### Part b)
# Explanation:
## Step1: Identify sides of $\angle 1$
$\angle 1$ is formed by two rays. From the diagram, the sides of $\angle 1$ are $BD$ and $BC$ (or $BB$? Wait, no, the rays are $BD$ and $BC$ with vertex $B$). Wait, looking at the diagram, $\angle 1$ has sides $BD$ and $BC$ (or maybe $BD$ and $BC$? Wait, the labels: $D$, $B$, $C$? Wait, the angle $\angle 1$: the two sides are the rays with endpoints at $B$, so one is $BD$ and the other is $BC$? Wait, no, maybe $BD$ and $BC$? Wait, the diagram has points $D$, $E$, $A$, $F$, $C$ with $B$ as the vertex. So $\angle 1$: sides are $BD$ and $BC$ (or $BB$? No, the rays are $BD$ (from $B$ to $D$) and $BC$ (from $B$ to $C$)).
# Answer:
The sides of $\angle 1$ are $\overrightarrow{BD}$ and $\overrightarrow{BC}$.

### Part c)
# Explanation:
## Step1: Find another name for $\angle 5$
$\angle 5$ has vertex $B$ and sides $BE$ and $BD$. So another name can be $\angle EBD$ (since the middle letter is the vertex, and the other two are the endpoints of the sides).
# Answer:
Another name for $\angle 5$ is $\angle EBD$.

### Part d)
# Explanation:
## Step1: Classify $\angle FBC$
Since $BF \perp AC$, $\angle FBC$ is a right angle (90 degrees), so it's a right angle.
## Step2: Classify $\angle EBF$
$\angle EBF$: let's see, if $BF \perp AC$, and $\angle EBF$: if we check the measure, but from the diagram, if $BF \perp AC$, then $\angle FBC = 90^\circ$, and $\angle EBF$: let's see, if $\angle EBF$ is greater than 90? Wait, no, maybe $\angle EBF$: if $BF \perp AC$, then $\angle FBA = 90^\circ$ (since $AC$ is a straight line? Wait, $AC$: $A$, $B$, $C$? Wait, $BF \perp AC$ means $\angle FBA$ and $\angle FBC$ are right angles (90 degrees). So $\angle EBF$: if $\angle EBF$ is, say, greater than 90? Wait, no, maybe $\angle EBF$ is obtuse? Wait, no, let's re - examine. Wait, $BF \perp AC$, so $\angle FBA = 90^\circ$. If $\angle EBF$ is, for example, if $\angle ABE$ is some angle, but maybe $\angle EBF$ is obtuse? Wait, no, maybe I made a mistake. Wait, $\angle FBC$: right angle (90°), $\angle EBF$: let's see, if $BF \perp AC$, then $\angle FBA = 90^\circ$. If $\angle EBF$ is, say, $\angle EBF=\angle EBA+\angle ABF$, but $\angle ABF = 90^\circ$, so if $\angle EBA$ is positive, then $\angle EBF>90^\circ$, so obtuse. $\angle ABC$: since $A$, $B$, $C$ are colinear (because $AC$ is a straight line, as $BF \perp AC$), so $\angle ABC$ is a straight angle (180°).
## Step1: $\angle FBC$
Since $BF\perp AC$, $m\angle FBC = 90^{\circ}$, so it is a right angle.
## Step2: $\angle EBF$
Assuming the diagram, if $BF\perp AC$ (so $\angle FBA = 90^{\circ}$) and $\angle EBF$ is greater than $90^{\circ}$ (since it includes $\angle FBA$ and another angle), so it is an obtuse angle.
## Step3: $\angle ABC$
Points $A$, $B$, $C$ are collinear (since $AC$ is a straight line with $B$ on it), so $m\angle ABC=180^{\circ}$, so it is a straight angle.
# Answer:
- $\angle FBC$: Right angle
- $\angle EBF$: Obtuse angle
- $\angle ABC$: Straight angle

### Part e)
# Explanation:
## Step1: Identify angle bisector
An angle bisector divides an angle into two equal parts. From the diagram, if $\angle 3=\angle 4$ (assuming, since $BF\perp AC$ and maybe $BA$ bisects some angle? Wait, no, looking at the diagram, if $BA$ divides an angle into two equal parts. Wait, the diagram has $\angle 3$ and $\angle 4$ with vertex $B$. If $BA$ is the bisector of $\angle FBE$ (since $BF\perp AC$, $\angle FBA = 90^{\circ}$, and if $\angle 3=\angle 4$, then $BA$ bisects $\angle FBE$). So the angle bisector is $\overrightarrow{BA}$ (or $BA$).
# Answer:
An angle bisector is $\overrightarrow{BA}$ (or $BA$).

### Part f)
# Explanation:
## Step1: Recall angle addition postulate
$\angle EBC=\angle EBD+\angle DBC$. Given $m\angle EBD = 36^{\circ}$ and $m\angle DBC = 108^{\circ}$.
## Step2: Calculate $m\angle EBC$
Using the angle addition postulate: $m\angle EBC=m\angle EBD + m\angle DBC=36^{\circ}+108^{\circ}=144^{\circ}$.
# Answer:
$m\angle EBC = 144^{\circ}$

### Part g)
# Explanation:
## Step1: Use perpendicularity
Since $BF\perp AC$, $m\angle FBA = 90^{\circ}$. Let $m\angle ABE=x$. Then $m\angle EBF=m\angle ABE + m\angle FBA$. Given $m\angle EBF = 117^{\circ}$ and $m\angle FBA = 90^{\circ}$.
## Step2: Solve for $m\angle ABE$
We have the equation $117^{\circ}=x + 90^{\circ}$. Subtract $90^{\circ}$ from both sides: $x=m\angle ABE=117^{\circ}-90^{\circ}=27^{\circ}$.
# Answer:
$m\angle ABE = 27^{\circ}$

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QUESTION IMAGE

Question

Explanation:

Part a)

Step1: Recall vertex definition

Step1: Identify sides of $\angle 1$

Step1: Find another name for $\angle 5$

Answer:

Part b)