A challenging geometry proof? 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.
OP's own answer shows that the key to the solution is to recognize that point $X$ lies on a hyperbola with foci $A$ and $C$ passing through $B$. Here's a "geometric" derivation of that fact.



Let the trisecting points of $\overline{BC}$ be $S$ and $T$. Let the angle bisector at $C$ meet $\overline{BX}$ at $D$, creating isosceles $\triangle BCD$. Let $\overleftrightarrow{DM}$ (with $M$ the midpoint of $\overline{BC}$ be the extended altitude of this triangle, and let $P$ be the projection of $X$ onto this line. 



Then we have
$$\left.\begin{align}
\text{Angle Bis. Thm} &amp;\implies \frac{|CX|}{|DX|}=\frac{|BC|}{|BD|} = \frac{2|BM|}{|BD|} \\[4pt]
\triangle DXP\sim\triangle DBM &amp;\implies \frac{|DX|}{|PX|}=\frac{|BD|}{|BM|}
\end{align}\right\}\implies
\frac{|CX|}{|PX|}=\frac{|CX|}{|DX|}\cdot\frac{|DX|}{|PX|}=2$$
Therefore, $\overleftrightarrow{DM}$ is the directrix, and $C$ the focus, of a hyperbola through $X$ with eccentricity $2$. 

Since trisection point $T$ divides $\overline{MC}$ in the ratio $1:2$, it must be a vertex of the hyperbola. Moreover, since $|ST|:|SC|=1:2$, it follows that $S$ is the center of the hyperbola. By symmetry across that center, $B$ and $A$ are the other vertex and focus, respectively, and the result follows. $\square$
Unfortunately, when I tried to edit DinosaurEgg's proof, the mods rejected it for "deviating from the original intent of the post".  With all due respect, I disagree with them, but it was a pretty dramatic revision.  So I'll post it here so that people can appreciate the elegance.  DinosaurEgg, if you see this, feel free to edit this into your answer if you like, and I'll delete this answer.



Construct $P$ on $\overline{AB}$ such that $\angle XPB\cong\angle B$. $\triangle XPB$ is obviously isosceles.  Since $\angle XCP=\theta$ and $\angle XPB=2\theta$, $\triangle XCP$ is also isosceles by the External Angle Theorem.  Therefore, $XB=XP=PC$. Dropping the perpendicular from $X$, it should be clear that $\cos B=\frac{PB/2}{XB}$, or $$PB=2XB\cos B$$ Since we know that $BC=3AC$, it follows that $AB=4AC$ and $PB=BC-PC=3AC-XB$.

Combining all of that with the Law of Cosines applied to $\triangle ABX$ yields

$$\begin{eqnarray*}
AX &amp; ^{2}= &amp; XB^{2}+AB^{2}-2AB\cdot XB\cos B\\
 &amp; = &amp; XB^{2}+AB^{2}-AB\cdot PB\\
 &amp; = &amp; XB^{2}+(4AC)^{2}-4AC(3AC-XB)\\
 &amp; = &amp; XB^{2}+16AC^{2}-12AC^{2}+4XB\cdot AC\\
 &amp; = &amp; XB^{2}+4XB\cdot AC+4AC^{2}\\
 &amp; = &amp; (XB+2AC)^{2}
\end{eqnarray*}$$