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In triangle \(ABC,M\) is the midpoint of \(\overline{AB}.\) Let \(D\) be the point on \(\overline{BC}\) such that \(\overline{AD}\) bisects \(\angle{BAC},\) and let the perpendicular bisector of \(\overline{AB}\)intersect \(\overline{AD}\) at \(E.\) If \(AB=44,AC=30,\) and \(ME=10,\) then find the area of triangle \(ACE\).

 

Here's what I got so far:

 

\(AE=2\sqrt{146}\)

If \(CD\) and \(BD\) have a "multiplication" of \(x\), then \(CD=15x\) and \(BD=22x\).

\([AME]=[BME]=110\)

That's all I know for sure. Help!

 

~ PikachuLovesKetchup

 Mar 27, 2022
 #1
avatar+1361 
+2

Alright, I'll take a shot...

 

Reflect \(\triangle AME\) over angle bisector \(\overline{AD}\).

 

You now have 2 triangles, and the sum of the areas equals \(\triangle ACE\)

 

The first triangle has a height of 10, and a base of 22 (Note: It has the same side lengths as \(\triangle AME\) because it is a reflection)

 

The second triangle has a height of 10, and a base of 8 (both bases sum to 30). 

 

Thus, the area is \(110 + 40 = \color {brown} \boxed {150}\)

 

Here is the messy diagram I made: 

 Mar 27, 2022
edited by BuilderBoi  Mar 27, 2022
 #2
avatar+117105 
+1

Deleted. Diagram was wrong.

It always helps if you read and inprete the question properly.

 

Thanks Builderboi.

I have only half checked what you have done it but it looks like a really nice solution.   laugh

 Mar 27, 2022
edited by Melody  Mar 27, 2022
edited by Melody  Mar 27, 2022
 #3
avatar+1361 
+2

Melody, your diagram has a mistake...

 

\(\overline{AB} = 44\), not  \(\overline{CB}\)

BuilderBoi  Mar 27, 2022
 #4
avatar+117105 
+1

Yes  thanks,

I found that when I was looking at what you did.    blush

Thanks for bringing it to my attention.  laugh

Melody  Mar 27, 2022

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