TITLE: A challenging geometry proof?
QUESTION [10 upvotes]: Given: $C$ on $\overline{AB}$ such that $BC=3AC$ and $m\angle B=2m\angle XCB$.
To show: $AX=2AC+BX$
I have verified this result with trigonometry and analytic geometry and double-checked my work with GeoGebra.  But it seems like such an elegant result that there should be a purely geometric proof.  Any ideas?
I was inspired to investigate this diagram trying to solve a different problem here on MSE.  As far as trying to come up with a proof on my own I tried constructing $M,N$ on $\overline{AX}$ such that $AM=AC$ and $NX=BX$ and drawing some isosceles triangles. That might be fruitful (since you'd only have to show $AM=MN$), but nothing leapt out at me quickly.

REPLY [2 votes]: Draw a circle with center at $X$ and radius $XB$ (look at the picture). 
Easy angle chase give $CE = EX (=BX=DX)$
Since the triangles $ADE$ and $ABD'$ are similar we have $$AD \cdot AD' = AE\cdot AB$$
so $$(AX-BX)(AX+BX) = (AC+BX)\cdot 4AC$$ 
so $$AX^2 = BX^2+4AC\cdot  BX+4AC^2 = (BX+2AC)^2$$
and we are done.