Solving $\tan x= \tan 2x\tan3x\tan4x$ and a related geometric problem My friend asked me a geometric problem.
In $\triangle ABC$, $\angle B=\angle C=70^\circ$. $D$ is an interior point of the triangle such that $\angle BCD=40^\circ$ and $\angle CBD=20^\circ$. Find $\angle BAD$.
If $\angle BAD=\theta$, it is not difficult to see that $\dfrac{\tan(40^\circ-\theta)}{\tan40^\circ}=\dfrac{\tan30^\circ}{\tan70^\circ}$ and I managed to show that $\tan(40^\circ-\theta)=\tan 10^\circ$ using trigonometric identities.
I actually proved that $\tan10^\circ\tan70^\circ=\tan30^\circ\tan40^\circ$, or equivalently, $\tan10^\circ=\tan20^\circ\tan30^\circ\tan40^\circ$. This result is so beautiful and make me interested in the equation $\tan x= \tan2x\tan3x\tan4x$, but I have difficulty in solving it. By plotting the graph, I can see that the solution is $180n^\circ$ or $60n^\circ\pm10^\circ$.
My questions are


*

*How to solve the original geometric problem without using trigonometry?

*How to solve the equation  $\tan x= \tan2x\tan3x\tan4x$?
Remark
Just find a solution to the second question a few minutes after posting it.  But I still want to see if there are other ways to solve it.
 A: Rewrite $\tan x= \tan2x\tan3x\tan4x$ as
$$\sin x \cos 2x\cos3x\cos 4x = \cos x \sin 2x\sin 3x\sin 4x$$
and factorize,
$$\sin x\cos 2x (\cos3x\cos 4x -4\cos^2 x\sin 3x\sin 2x)=0$$
Further factorize with $\cos 3x = \cos x(2\cos 2x -1)$, 
$$\sin x\cos 2x \cos x [(2\cos 2x -1)\cos 4x -4\cos x\sin 3x\sin 2x)]=0\tag 1$$
Recognize $\cos x \ne 0$, $\cos 2x \ne 0$ and
$$(2\cos 2x -1)\cos 4x =\cos2x+\cos6x-\cos4x$$
$$4\cos x\sin 3x\sin 2x=2(\sin4x+\sin2x)\sin2x= \cos2x-\cos6x+1-\cos4x$$
to reduce the equation (1) to,
$$\sin x(2\cos 6x -1)=0 $$
which leads to $\sin x =0$ and $\cos6x=\frac12$. Thus, the solutions are
$$x=n\pi,\>\>\>\>\> x = \frac{n\pi}3\pm\frac\pi{18}$$
A: 2) If $\tan3x=0$,  so we can check it easily.  
Let $\tan3x\neq0.$ 
Thus, we need to solve:
$$\tan{x}\cot3x+1=\tan2x\tan4x+1$$ or
$$\frac{\sin4x}{\cos{x}\sin3x}=\frac{\cos2x}{\cos2x\cos4x}$$ or
$$\sin4x\cos4x=\sin3x\cos{x}$$ or
$$\sin8x=\sin4x+\sin2x$$ or
$$\sin8x-\sin4x=\sin2x$$ or
$$\sin2x(2\cos6x-1)=0.$$ Can you end it now?
The first problem.
Take $\Delta BFG$ such that $BG=GF$ and $\measuredangle G=20^{\circ}.$
Let $K\in GF$, $M\in GK$ and $E\in GB$ such that $BF=BK=EK=EM.$
Thus, $\measuredangle KBF=20^{\circ}$ and
$$\measuredangle EBK=80^{\circ}-20^{\circ}=60^{\circ},$$ which gives
$$BE=BK=EK=EM.$$
Also, $$\measuredangle EMK=\measuredangle EKM=180^{\circ}-\measuredangle EKB-\measuredangle BKF=180^{\circ}-60^{\circ}-80^{\circ}=40^{\circ}$$ and since $\measuredangle G=20^{\circ},$ we obtain: $$\measuredangle GEM=40^{\circ}-20^{\circ}=20^{\circ},$$ which gives
$$GM=ME=EK=BK=EB=BF.$$
Thus, $$\measuredangle EBM=\measuredangle EMB=\frac{1}{2}\measuredangle GEM=10^{\circ},$$
which gives $$\measuredangle MBF=80^{\circ}-10^{\circ}=70^{\circ},$$
$$\measuredangle MBK=50^{\circ},$$
$$\measuredangle KBF=20^{\circ}.$$
Also, we have:
$$\measuredangle BKM=60^{\circ}+40^{\circ}=100^{\circ}$$ and $$\measuredangle BMK=40^{\circ}-10^{\circ}=30^{\circ}.$$
Now, let $EK\cap BF=\{C\}$ and $N$ be placed on the line $BC$ such that $B$ is a mid-point of $NF$. 
But $BN=BE$ and $$\measuredangle NBE=180^{\circ}-80^{\circ}=100^{\circ}=\measuredangle MEK,$$ which gives $$\Delta NBE\cong\Delta MEK,$$ which says $$NE=MK.$$
In another hand, $$\measuredangle ECN=180^{\circ}-\measuredangle N-\measuredangle NEC=180^{\circ}-40^{\circ}-100^{\circ}=40^{\circ},$$
which gives $$EC=NK=MK.$$
Thus, $$\Delta MEC\cong\Delta BKM,$$ which gives $$MC=BM$$ and
$$\measuredangle MCB=\measuredangle MBC=70^{\circ}.$$
Id est, $$\Delta MBC\cong\Delta ABC,$$ which gives $M\equiv A$  and since $\measuredangle KCB=40^{\circ},$ we obtain: $K\equiv D,$ which says
$$\measuredangle BAD=\measuredangle BMK=30^{\circ}.$$
A: Geometric solution

Construct the point $E$ as a reflection of the point $D$ w.r.t. the vertical line through $A$,
$\angle DAE=\theta-(40^\circ-\theta)=2\theta-40^\circ$.
Point $H=BD\cap CE$.
$\triangle BDE$, $\triangle CDE$ and $\triangle EHD$ are isosceles, 
$\angle EBD=\angle BDE=\angle DEC=\angle ECD=20^\circ$, $|BE|=|CD|=|DE|$.
Point $D_1:DD_1\perp AC,\ |DD_1|=|DE|$, point $F=AC\cap DD_1$.
From $\triangle CDF$, $|DF|=|D_1F|=\tfrac12\,|CD|=\tfrac12\,|DE|$.
Similarly,
point $E_1:DD_1\perp AB,\ |EE_1|=|DE|$, point $G=AB\cap EE_1$.
From $\triangle BEG$, $|EG|=|E_1G|=\tfrac12\,|BE|=\tfrac12\,|DE|$.
Noe we have $\angle D_1AD=\angle EAE_1=\angle DAE$,
$\angle FAD=\angle EAG=\tfrac12\,\angle DAE$, 
so
\begin{align}
2\angle DAE&=\angle CAB
,\\
2(2\theta-40^\circ)&=40^\circ
,\\
\theta&=30^\circ
.
\end{align} 
A: There are two nice solutions to my second question. Here I would like to share mine.  I am not intended to answer my own question and I have yet to solve the first problem.
When $x,2x,3x,4x\notin\{(n+\frac12) \pi:n\in\mathbb{Z}\}$, we have
\begin{align*}
\sin x\cos 4x \cos 2x\cos 3x&=\cos x\sin 4x \sin2x\sin3x\\
\frac12(\sin5x-\sin3x)\cdot\frac12(\cos x+\cos 5x)&=\frac12(\sin5x+\sin3x)\cdot\frac12(\cos x-\cos5x)\\
\sin5x\cos5x-\sin3x\cos x&=0\\
\sin10x-\sin4x-\sin2x&=0\\
2\cos6x\sin4x-\sin4x&=0\\
\sin4x(2\cos6x-1)&=0
\end{align*}
So, $\displaystyle x=\frac{n\pi}{4}$ or $\displaystyle \frac{n\pi}3\pm\frac{\pi}{18}$.
As $x,2x,3x,4x\notin\{(n+\frac12) \pi:n\in\mathbb{Z}\}$, we have $\displaystyle x=n\pi$ or $\displaystyle \frac{n\pi}3\pm\frac{\pi}{18}$.
