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.


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 $\angle EAD = 3x$. 
