+0

# Help! I'm totally stuck, and I don't know how to continue the diagram.

0
74
4
+32

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
+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
+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.

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

Melody, your diagram has a mistake...

$$\overline{AB} = 44$$, not  $$\overline{CB}$$

BuilderBoi  Mar 27, 2022
#4
+117105
+1

Yes  thanks,

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

Thanks for bringing it to my attention.

Melody  Mar 27, 2022