$ABCD$ is a convex quadrilateral. If $\angle BAC=10°$, $\angle CAD=40°$, $\angle ADB=50°$, $\angle BDC=20°$, then find $\angle CBD$. 
$ABCD$ is a convex quadrilateral. If $\angle BAC=10°$, $\angle CAD=40°$, $\angle ADB=50°$, $\angle BDC=20°$, then find $\angle CBD$.

The problem appeared in a local math contest last month. I know these kinds of angle chasing problems are called adventitious angles which is discussed in this wikipedia article. But as this is not a $80$-$80$-$20$ triangle, I think this is a different problem and I couldn't solve the problem completely. To understand this kind of angle chasing problems, I tried to solve the problem both in trigonometric and geometric way. Here are my workings to do that:
Trigonometric solution:

Let $P$ be the intersection of $AC$ and $BD$. Then, applying sine rule in $\triangle PAB$, $\triangle PBC$, $\triangle PCD$ and $\triangle PAD$ and multiplying them we have $$\sin(x)\sin(70°)\sin(50°)\sin(10°)=\sin(90-x)\sin(20°)\sin(40°)\sin(80°)$$ $$\implies \tan(x)\tan(60°+10°)\tan(60°-10°)\tan(10°)=1.$$
Using calculator, I found $x=60°$. But as the contest doesn't allow using calculators, I think the equation can't be solved without knowing the value of $\tan(10°)$. So, I need a trigonometric solution that doesn't use calculators and doesn't use the value of $\tan(10°)$.
Geometric solution:
To solve the problem geometrically, I noticed that $\triangle ABD$ and $\triangle ACD$ are both isosceles with $AB=BD$ and $AC=AD$ respectively. I tried to make an equilateral triangle then. But I don't know how to do that.
I also tried drawing circles with center $B$ and arc $AD$. But $C$ doesn't lie on the circle as $\angle ACD=70°$. So, I couldn't proceed further. And same for circle with center $A$ and arc $CD$.

So, I need to complete my solutions. Any helpful idea is welcome.
 A: 
Start off by constructing an equilateral triangle $AMD$ as shown in the diagram above. Note that since $AB=BD$, then $\sphericalangle AMB=\sphericalangle BMD=30^\circ$ by symmetry. We easily calculate $\sphericalangle MAB=10^\circ$ which, since $AM=AC$, implies $\triangle AMB\cong\triangle ABC$. Thus, $\sphericalangle AMB=\sphericalangle ACB=30^\circ$ and so the required angle is indeed equal to $60^\circ$.
A: Theorem.
$$\tan (x + \pi/3) \tan (x - \pi/3) \tan x = -\tan 3x.$$
Proof.  Let $y = \tan x$.  Then
$$\tan (x \pm \pi/3) = \frac{\tan x \pm \tan \pi/3}{1 \mp \tan x \tan \pi/3} = \frac{y \pm \sqrt{3}}{1 \mp y \sqrt{3}},$$ and $$\tan 2x = \frac{2 \tan x}{1 - \tan x^2} = \frac{2y}{1-y^2}.$$  Then
$$\begin{align}
\tan 3x &= \frac{\tan x + \tan 2x}{1 - \tan x \tan 2x} \\
&= \frac{y + \frac{2y}{1-y^2}}{1 - \frac{2y^2}{1-y^2}} \\
&= \frac{y(3-y^2)}{1-3y^2} \\
&= -\frac{(y + \sqrt{3})(y - \sqrt{3})}{(1-y\sqrt{3})(1+y\sqrt{3})} y \\
&= -\tan (x+\pi/3) \tan (x-\pi/3) \tan x.
\end{align}$$
This is of course not the simplest approach, but it satisfies the criteria you specified for a trigonometric proof.
A: We  use the result
$$\sin x \sin\left(\frac{\pi}{3}+x \right)\sin \left(\frac{\pi}{3}-x \right) =\frac{1}{4}\sin 3x$$
Using the law of sines in triangles $CPD,PDA,PAB,BPC$, we end up with
$$\tan x=\frac{\sin\left(\frac{\pi}{9} \right) \sin\left(\frac{2\pi}{9} \right) \sin \left(\frac{4\pi}{9} \right)}{ \sin\left(\frac{\pi}{18} \right) \sin\left(\frac{5\pi}{18} \right) \sin \left(\frac{7\pi}{18} \right)}$$
$$=\frac{\sin\left(\frac{\pi}{3} \right)}{\sin \left(\frac{\pi}{6} \right)}=\sqrt{3}$$
From which, it follows that $x=\frac{\pi}{3}$, assuming convexity
