Angles inside an equilateral triangle 
We have a point $P$ inside the equilateral triangle $\triangle ABC$, such that $\angle PAC = x$, $\angle PCA = 3x$ and $\angle PBC = 2x$. Find the value of $x$ in degrees.
I solved the problem in GeoGebra, and I know the value for $x$ is $6°$. Also I solved this problem by using the Trigonometry Ceva's Theorem:
$$
\sin(60°-2x)\sin(x)\sin(60°-3x)=\sin(2x)\sin(60°-x)\sin(3x)
$$
Where I used WolframAlpha to get $x=6°$.
However, I'm looking for a geometric solution. This is what I have reacher so far.

First, draw the circle that lies over $B$, $P$ and $C$. Then extend sides $AB$ and $AC$ to get point $E$ and $D$ (we have another equilateral triangle $\triangle AED$). Due to properties of angles inscribed in a circle $\angle PDC = 2x$ and $\angle PED = 3x$. Then trace the bisector $DG$, we get that $\angle GDP=30°-2x$ and $\angle PGD = 3x$. Here I got stuck. I think, I should proof that $GP = PD$, but I don't know how.
I would appreciate any contribution to this solution, or if you have a different approach and solution I would glad to hear it.
 A: 
Considering figure(left one) in kite ABFC we have:
$\angle BFC+2(\angle ACF)+\angle BAC=(4-2)\times 180=360$
Or:
$(120-2x)+2(\angle ACF)+60=360\rightarrow 180=2(\angle ACF-x)\rightarrow \widehat{ACF}=90+x$
Also in kite ABCG we have:
$(\angle AGC=8x)+60+2(\angle BCG)=360\rightarrow 2(\angle BCG) +8x=300\rightarrow \angle BCG=150-4x$
$BCF-ACF=150-4x-90 -x=60-5x$
$\Rightarrow \angle ACG-\angle BCF=60-5x\Rightarrow 60-5x=ACG-(30+x)$
$\Rightarrow \angle ACG=90-4x$
This equation has following solutions:
$(\angle ACG, x)= (66,6), (70, 5), (74, 4)\cdot\cdot\cdot$
Only $x=6^o$ is competent with equilateral triangle. Figure on the right shows when $x=5$ for example.
A: For a geometric solution, make a circle with center $B$ and radius $BA$, thus passing through vertex $C$ of the equilateral triangle.
Let $A$, $D$ be vertices of the regular pentagon in that circle, and $A$, $E$ vertices of the regular decagon. Join $BD$, $BE$. Also join $AD$ and $CE$, intersecting at $P$, and draw $BP$ through to $F$..

Since $\angle DBC=72-60=12^o$, and $\angle BDG=\frac{180-36}{2}=72^o$, then $BC\perp GD$ and $$\angle PBC=\angle DBC=12^o$$and at the circumference$$\angle PAC=6^o$$And since in the decagon $\angle EBA=36^o$, then at the circumference$$\angle PCA=18^o$$
Hence we have$$\angle PBC=2\angle PAC$$and$$\angle PCA=3\angle PAC$$
