+40 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
+2666 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: 
+118609 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. 
+2666 Melody, your diagram has a mistake...
\(\overline{AB} = 44\), not \(\overline{CB}\)
