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enter image description here 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.

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  • $\begingroup$ You should provide at least a sketch of your proof. It could save answerers from wasting time duplicating your effort. Perhaps someone can even "see" a geometric approach in your equations. $\endgroup$ – Blue Aug 16 '19 at 22:01
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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.

enter image description here

Then we have $$\left.\begin{align} \text{Angle Bis. Thm} &\implies \frac{|CX|}{|DX|}=\frac{|BC|}{|BD|} = \frac{2|BM|}{|BD|} \\[4pt] \triangle DXP\sim\triangle DBM &\implies \frac{|DX|}{|PX|}=\frac{|BD|}{|BM|} \end{align}\right\}\implies \frac{|CX|}{|PX|}=\frac{|CX|}{|DX|}\cdot\frac{|DX|}{|PX|}=2$$ 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. $\square$

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  • 1
    $\begingroup$ That's really nice. I've never seen a use for the directrix and eccentricity of a hyperbola before, so it's very reassuring to see applications for it! $\endgroup$ – Matthew Daly Aug 17 '19 at 0:11
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    $\begingroup$ Thanks. :) I'm always pleased when the focus-directrix definition of conics (not just parabolas) comes into play. ... I was right about how posting your own proof would be a time-saver. Even if my final argument hadn't used the hyperbola, just knowing it was there made constructing a GeoGebra sketch super-simple, freeing me to get on with exploring possible approaches. So, thanks for that, too. :) $\endgroup$ – Blue Aug 17 '19 at 0:23
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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 $\overline{AB}$ we see that $$\tan\theta=\frac{y}{x}$$ $$\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}=\frac{y}{1-x}$$

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

$$\frac{y}{1-x}=\frac{2y/x}{1-y^2/x^2}=\frac{2xy}{x^2-y^2}$$ $$2x(1-x)=x^2-y^2$$ $$y^2=x^2-2x(1-x)$$ $$y^2=3x^2-2x$$

This is the equation of a hyperbola with foci at $(-\frac{1}{3},0)=A$ and $(1,0)=B$ and a vertex at $(\frac{2}{3},0)$. Since this hyperbola is the locus of points $X$ such that $AX-BX=\frac{2}{3}=2AC$, the statement follows.

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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.

enter image description here

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  • $\begingroup$ Huh. For some reason, I think of the power of the point as a collegiate geometry topic that I don't understand. It might be that I actually don't, but when people describe the secant-secant theorem (which is very much a high school geometry fact) as the power of the point, I just assume that I won't understand it. Anyways, yes! It's curious that DinosaurEgg got the same essential fact from the Law of Cosines instead. $\endgroup$ – Matthew Daly Oct 24 '19 at 20:22
  • $\begingroup$ No power of the point now? $\endgroup$ – Aqua Oct 27 '19 at 20:04
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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

$$BP=2BX\cos2\theta~~,~~XC=2BX\cos\theta$$

and since $BP+PC=3x=BX(1+2\cos2\theta)$ ($AB=4x$ as per the sketch provided) we express the following lengths in terms of $x, \theta$: $$BX=\frac{3x}{1+2\cos2\theta}~~~,~~~ BP=\frac{6x\cos2\theta}{1+2\cos2\theta}~~~,~~~CX=\frac{6x\cos\theta}{1+2\cos2\theta}$$

The law of cosines on $ABX$ reads:

$$\begin{align}AX^2=&AB^2+BX^2-2AB\cdot BX\cos2\theta\\=&\frac{x^2}{(1+2\cos2\theta)^2}\Big(16(1+2\cos2\theta)^2-24(1+2\cos2\theta)\cos2\theta+9\Big)\\=&\frac{x^2}{(1+2\cos2\theta)^2}\Big(25+40 \cos2\theta+16 \cos^2 2\theta\Big)\\=&\frac{x^2}{(1+2\cos2\theta)^2}(2(1+2\cos2\theta)+3)^2\\=&(BX+2AC)^2\end{align}$$

and the proof is complete.

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  • $\begingroup$ Edited and corrected. As pointed out, the angle is unconstrained. The equality naturally comes out by simple algebraic manipulations. $\endgroup$ – DinosaurEgg Aug 17 '19 at 2:18
  • $\begingroup$ Also, I think since trigonometric manipulations did not play any deep role here (they only serve as mere labels for ratios between lengths if you think about it),my intuition is that there must be a solution doesn't make use of the law of cosines at all. $\endgroup$ – DinosaurEgg Aug 17 '19 at 2:43
  • $\begingroup$ I think you need the relationship between $BX$ and $PX$ (although it follows more simply without LoC), and also the last LoC brings $AX$ into the picture. But the proof could be shorter -- you definitely use that XPC is isosceles, but the calculation for $CX$ is extraneous. I'll play with it and propose an edit (plus a diagram, which will help people). But wow! This either avoids (or hides very well) the hyperbola connection that Blue and I relied on. Fantastic work! $\endgroup$ – Matthew Daly Aug 17 '19 at 9:56
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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.

enter image description here

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

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

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