Ttdfjt

Given: $C$ on $overlineAB$ such that $BC=3AC$ and $mangle B=2mangle XCB$.

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 $overlineAX$ 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 $overlineBC$ be $S$ and $T$. Let the angle bisector at $C$ meet $overlineBX$ at $D$, creating isosceles $triangle BCD$. Let $overleftrightarrowDM$ (with $M$ the midpoint of $overlineBC$ be the extended altitude of this triangle, and let $P$ be the projection of $X$ onto this line.

Then we have
$$left.beginalign
textAngle Bis. Thm &implies fracCX=frac = frac2 \[4pt]
triangle DXPsimtriangle DBM &implies frac=frac
endalignright}implies
fracCX=fracCXcdotfrac=2$$
Therefore, $overleftrightarrowDM$ is the directrix, and $C$ the focus, of a hyperbola through $X$ with eccentricity $2$.

Since trisection point $T$ divides $overlineMC$ 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. $square$

Here is my non-Euclidean proof of the fact, as it was suggested that it might inspire people. Understand that I want a Euclidean proof of the statement, so this proof is not what I'm looking for here.

Arrange the diagram on the Cartesian plane such that C is at the origin and B is at (1,0). Let $(x,y)$ be the coordinates of point X. Then, dropping a perpendicular from X to $overlineAB$ we see that
$$tantheta=fracyx$$
$$tan2theta=frac2tantheta1-tan^2theta=fracy1-x$$

by the double angle formula. Combining those two formulas gives us

$$fracy1-x=frac2y/x1-y^2/x^2=frac2xyx^2-y^2$$
$$2x(1-x)=x^2-y^2$$
$$y^2=x^2-2x(1-x)$$
$$y^2=3x^2-2x$$

This is the equation of a hyperbola with foci at $(-frac13,0)=A$ and $(1,0)=B$ and a vertex at $(frac23,0)$. Since this hyperbola is the locus of points $X$ such that $AX-BX=frac23=2AC$, the statement follows.

I hope that this proof is the geometry you're looking for (it is geometry I was taught in high school at least).

Let $P$ be the point such that the line $PX$ forms the angle $angle BPX=angle PBX$. This should lie between the lines $XC$ and $XB$. The triangles $PBX$ and $XPC$ are isosceles, and therefore we have $XB=XP=PC$. Applying the law of cosines for both triangles we obtain that

$$BP=2BXcos2theta~~,~~XC=2BXcostheta$$

and since $BP+PC=3x=BX(1+2cos2theta)$ ($AB=4x$ as per the sketch provided) we express the following lengths in terms of $x, theta$:
$$BX=frac3x1+2cos2theta~~~,~~~ BP=frac6xcos2theta1+2cos2theta~~~,~~~CX=frac6xcostheta1+2cos2theta$$

The law of cosines on $ABX$ reads:

$$beginalignAX^2=&AB^2+BX^2-2ABcdot BXcos2theta\=&fracx^2(1+2cos2theta)^2Big(16(1+2cos2theta)^2-24(1+2cos2theta)cos2theta+9Big)\=&fracx^2(1+2cos2theta)^2Big(25+40 cos2theta+16 cos^2 2thetaBig)\=&fracx^2(1+2cos2theta)^2(2(1+2cos2theta)+3)^2\=&(BX+2AC)^2endalign$$

and the proof is complete.

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 $overlineAB$ such that $angle XPBcongangle B$. $triangle XPB$ is obviously isosceles. Since $angle XCP=theta$ and $angle XPB=2theta$, $triangle XCP$ is also isosceles by the External Angle Theorem. Therefore, $XB=XP=PC$. Dropping the perpendicular from $X$, it should be clear that $cos B=fracPB/2XB$, or $$PB=2XBcos B$$ 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

$$begineqnarray*
AX & ^2= & XB^2+AB^2-2ABcdot XBcos B\
& = & XB^2+AB^2-ABcdot PB\
& = & XB^2+(4AC)^2-4AC(3AC-XB)\
& = & XB^2+16AC^2-12AC^2+4XBcdot AC\
& = & XB^2+4XBcdot AC+4AC^2\
& = & (XB+2AC)^2
endeqnarray*$$