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

PikachuLovesKetchup Mar 27, 2022

#1**+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:

BuilderBoi Mar 27, 2022

#2**+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.

Melody Mar 27, 2022

#3**+2 **

Melody, your diagram has a mistake...

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

BuilderBoi
Mar 27, 2022