Bisecting and Trisecting Angles: Problem

RoseOfStone241

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Sep 5, 2006
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In my book this is the information given:

Angle BAC = 120 degrees, and points D, E, F are the interior of BAC. Ray AD bisects angle BAF, and ray AE bisects angle CAF. Find m of angle DAE.

Alright, so i think that's the best way to describe the diagram. BAC is obtuse and rays AD, AF, and AE are drawn from BAC's vertex of A.

Obviously BAD and DAF are congruent, and FAE and EAC are congruent.

Other than this info., I'm lost on where to begin.

Any help would be much appreciated ^-^
 
Bisects means half the measure.
\(\displaystyle m\left( {\angle BAF} \right) = \frac{{m\left( {\angle BAC} \right)}}{2}.\)

\(\displaystyle m\left( {\angle BAD} \right) = \frac{{m\left( {\angle BAF} \right)}}{2}.\)

Now you finish it.
 
Bisecting And Assumptions

I understand that, but we cannot assume that ray AF bisects angle BAC.

I also understand that BAD is half of BAF. ^-^
 
ray AD bisects angle BAF ...
BAD = DAF = "a"

ray AE bisects angle CAF ...
FAE = EAC = "b"

2a + 2b = 120
a + b = 60

now ... DAE = DAF + FAE = a + b
 
How does that help? lol.

I understand everything you said skeeter. I even did the algebra myself. But I'm still lost. :?
 
\(\displaystyle \[
\begin{array}{l}
2m(\angle DAF) = m(\angle BAF) \\
2m(\angle FAE) = m(\angle FAC) \\
2m(\angle DAE) = m(\angle BAF) + m(\angle FAC) \\
\\
\end{array}\)
 
Alright, maybe i'm too tired, or I'm just missing something here.

Aren't we just re-establishing what i've already written?

I can tell all that info. from the diagram...
 
Just think about it.

\(\displaystyle m(\angle BAF) + m(\angle FAC) = m(\angle BAC) = 120\)
 
Yes, but BAF and FAC are different measurements. They're not bisected.

I see what you're saying, but ray AF doesn't bisect BAC.

Maybe I'm just missing this. It's certainly not clear to me at all. Can one of you explain it a bit rather than give me equations please? :)
 
GOOD Lord! I asked you to think!

\(\displaystyle \begin{array}{l}
2m(\angle DAE) = m(\angle BAF) + m(\angle FAC) \\
m(\angle BAF) + m(\angle FAC) = m(\angle BAC) = 120 \\
m(\angle DAE) = 60 \\
\end{array}\)
 
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