Geometric proof for $\tan{(3x)}=\tan{(x)}\tan{\left(\frac{\pi}{3}-x\right)}\tan{\left(\frac{\pi}{3}+x\right)}$ Are there geometric proofs for the identitity
$$\tan{(3x)}=\tan{(x)}\tan{\left(\frac{\pi}{3}-x\right)}\tan{\left(\frac{\pi}{3}+x\right)}$$
My try:
Thank in advances.
 A: 
I have another simple proof. 
Assume $\angle AFD = \frac{\pi}{3}$, $\angle FAD = 4x$, and $\angle FDA = 4y$. Obviously, $y = \frac{\pi}{6}-x$, $\textit{i.e.}$, $x+y = \frac{\pi}{6}$.
$AQ$ bisects $\angle FAD$, and $AC$ bisects $\angle FAQ$. 
Also, $DP$ bisects $\angle FDA$, and $DB$ bisects $\angle FDP$. 
And $R = AQ \cap DP$, $H = AQ \cap DB$, $G= AC \cap DP$, $E = AC \cap DB$. 
Then, $\angle AED = \angle AFD + \angle FAE + \angle FDE = 
\frac{\pi}{3} + x + y = \frac{\pi}{2}$. Similarly, $\angle ARD = \frac{2\pi}{3}$. 
$\text{area of }  \triangle ADP  =\frac{1}{2} AR \cdot DP \sin \frac{2\pi}{3}
= \frac{1}{2} AP \cdot DF \sin \frac{\pi}{3}$.
Thus, $AR\cdot DP = AP \cdot DF$. 
It means that $AR:AP = DF:DP$. 
We know that $AR:AP=RG:GP$, and $DF:DP=FB:BP$. Thus $RG:GP = FB:BP$. Therefore $RF \parallel GB$. 
Because $R$ is the incenter of $\triangle AFD$, $\angle RFP = \frac{\pi}{6}$.
Hence, $\angle GBP = \frac{\pi}{6}$. 
Thus, $\angle BGE = \angle BAG + \angle ABG = \frac{\pi}{6} + x$, and 
$\angle EBC = \angle EBG = \frac{\pi}{2} - \angle BGE = \frac{\pi}{3} - x$. 
Additionally, $\angle ECD = \frac{\pi}{2} - y = \frac{\pi}{2} - \left(\frac{\pi}{6} -x \right) = \frac{\pi}{3} + x$. Therefore, $\square ABCD$ satisfies the condition of problem, and we showed that $\angle EAD = 3x$. 
A: This is not a geometric proof (the one I mentioned in the comments still eludes me) but I hope you find it interesting nonetheless. By the Fourier series for $\log\sin$ and $\log\cos$
$$ \log\tan x = -2\sum_{k\geq 0}\frac{\cos((4k+2)x)}{2k+1} = -2\,\text{Re}\sum_{k\geq 0}\frac{e^{(2k+1)2ix}}{2k+1}  \tag{1}$$
holds for any $x\in\left(0,\frac{\pi}{2}\right)$, so the given identity is a simple consequence of the discrete Fourier transform, $\mathbb{1}_{\equiv 0\!\!\pmod{3}}(n)=\frac{1+\omega^n+\omega^{2n}}{3}$.
A: 
$A,B,C,D,E$ are the same points in the problem. 
$F= AB \cap CD$. 
$G$ and $C$ are symmetric with respect to $E$. $H$ and $B$ are symmetric with respect to $E$.
$P = GD \cap AF$, $Q = AH \cap DF$. $R = GD \cap AH$. 
(In my figure, I forgot to write $R$. Sorry.)
We want to show that $R$ is the incenter of $\triangle ADF$.
because $\angle AFD = \angle AED - \angle FAC - \angle FDB$, 
$\angle AFD = \frac{\pi}{3}$.
$\angle EBC = \angle EBG = \angle EHC = \frac{\pi}{3} - x$. 
and $\angle ECB = \angle EGB = \angle ECH = \frac{\pi}{6} + x$.
Because $\angle ABG = \angle EGB - \angle GAB$, 
$\angle ABG = \frac{\pi}{6}$.
Also, because $\angle HDC = \frac{\pi}{6} -x$, and $\angle DCH = \angle EHC - \angle HDC$, 
$\angle DCH = \frac{\pi}{6}$.
$\triangle DCH \equiv \triangle DGH$, and $\triangle ABG \equiv \triangle AHG$.
Thus, $\angle AHG = \angle DGH = \frac{\pi}{6}$. Thus, $RG=RH$.
Additionally, $\angle PRQ = \pi - \angle RGH - \angle RHG = \frac{2\pi}{3}$, 
and $\angle PRQ + \angle PFQ = \pi$. Hence, four points $R, P, F, Q$ are on the same circle. 
Because $\square BCHG$ is a rhombus, let $BG=CH=l$.
$\angle BPG = \pi - \angle AFD - \angle FDP = \frac{\pi}{3}+2x$.
In $\triangle BPG$, 
$$\frac{GP}{\sin\frac{\pi}{6}} = \frac{l}{\sin(\frac{\pi}{3}+2x)}$$
Similarly, $\angle CQH = \pi - \angle AFD - \angle FAQ = \frac{2\pi}{3}-2x$.
In $\triangle CQH$, 
$$\frac{HQ}{\sin\frac{\pi}{6}} = \frac{l}{\sin(\frac{2\pi}{3}-2x)}$$
$\sin(\frac{\pi}{3}+2x) = \sin(\frac{2\pi}{3}-2x)$, thus $GP=HQ$. 
Hence, $RP = RG + GP = RH + HQ = RQ$. 
Because $RP=RQ$ and $R,P,F,Q$ are on the same circle, $\angle RFP = \angle RFQ$.
Thus, we know $\angle RAF = 2x$, $\angle RFA = \angle RFD = \frac{\pi}{6}$,
$\angle RDF = \frac{\pi}{3}-2x$, and $\angle RAD + \angle RDA = \frac{\pi}{3}$.
Finally, we use the trigonometric form of Ceva's theorem. 
$$\frac{\sin{\angle RAD}}{\sin{\angle RAF}} \times \frac{\sin{\angle RDF}}{\sin{\angle RDA}} \times \frac{\sin{\angle RFA}}{\sin{\angle RFD}} = 1$$
It is equal to 
$$ \frac{\sin{\angle RAD}}{\sin 2x} \times \frac{\sin{\frac{\pi}{3}-2x}}
{\sin{\frac{\pi}{3}- \angle RAD}} = 1 $$
$\angle RAD$ should be $2x$. Because if $\angle RAD < 2x$, then the left-hand side of the above equation is less than $1$, 
and if $\angle RAD > 2x$, then the left-hand side of the above equation is greater than $1$.
Thus $R$ is the incenter of $\triangle FAD$, and $\angle EAD = 3x$. 
A: I do not know if there is a simpler geometrical incarnation, however this triplication formula (or, better, the analogous triplication formula for the sinus) is at the heart of the trigonometric proof of the celebrated Morley's trisector theorem.
