Simplifiying the trigonometry equation to show that triangle is isosceles In a triangle $ABC$ , $X$ is a point inside it , its given that Angle $BAX = 10^\circ$ , Angle $ABX = 20^\circ$ , Angle $XAC = 40^\circ$ and Angle $XCA = 30^\circ$, prove that triangle is isosceles.

My method was using trigonometry to arrive at value of the unknown angles, and then showing its isoceles. Lets call Angle $XCB$ to be $\alpha$, then applying sin rule on triangle $ABX$ and $AXC$, and similarly for other two pairs keeping in mind one side is common to them, we get arrive at $\sin(\alpha)\sin 20^\circ\sin40^\circ  = \sin(80^\circ -\alpha) \sin 10^\circ \sin 30^\circ$. Now my question is how do we solve such trigonometric equation in a restricted domain to show that only solution to this is $20^\circ$? Implying $AB = BC$.

 A: 
Comment: In figure AF, CG and BH are altitudes of triangle.It can be shown by angle chasing that:
$\overset{\large\frown}{ED}=20^o$
$\overset{\large\frown}{DH}=40^o$
$\Rightarrow \angle DBH=20^o$
These points must be shown:

*

*X is on altitude AF.


*$\widehat{GCB}=10^o$
$\angle GBH=50^o$
$\Rightarrow\overset{\large\frown}{GH}=100^o$
$\triangle AFB\sim\triangle CBG$
These two triangles are symmetric about BH, that is BH bisect angle ABC; it also bisect segment AC which results in $AH=HC$ and $BA=BC$ and triangle ABC is isosceles.
A: Here is an approach using the Law of Sines and the Law of Cosines, that is not too bad.

$$
\begin{align}
BX
&=AB\,\frac{\sin\left(10^{\large\circ}\right)}{\sin\left(150^{\large\circ}\right)}\tag{1a}\\[6pt]
&=2AB\sin\left(10^{\large\circ}\right)\tag{1b}\\[6pt]
CX
&=AX\,\frac{\sin\left(40^{\large\circ}\right)}{\sin\left(30^{\large\circ}\right)}\tag{2a}\\
&=AB\,\frac{\sin\left(20^{\large\circ}\right)}{\sin\left(150^{\large\circ}\right)}\,\frac{\sin\left(40^{\large\circ}\right)}{\sin\left(30^{\large\circ}\right)}\tag{2b}\\[6pt]
&=4AB\sin\left(20^{\large\circ}\right)\sin\left(40^{\large\circ}\right)\tag{2c}\\[12pt]
BC^2
&=BX^2+CX^2-2BX\,CX\,\cos\left(100^{\large\circ}\right)\tag{3a}\\[6pt]
&=AB^2\left(4\sin^2\left(10^{\large\circ}\right)+16\sin^2\left(20^{\large\circ}\right)\sin^2\left(40^{\large\circ}\right)\right.\\
&\phantom{=AB^2\left(\right.}\left.{}-16\sin\left(10^{\large\circ}\right)\sin\left(20^{\large\circ}\right)\sin\left(40^{\large\circ}\right)\cos\left(100^{\large\circ}\right)\right)\tag{3b}\\[6pt]
&=AB^2\left(\color{#C00}{4\sin^2\left(10^{\large\circ}\right)}+\color{#090}{16\sin^2\left(20^{\large\circ}\right)\sin^2\left(40^{\large\circ}\right)}\right.\\
&\phantom{=AB^2\left(\right.}\left.{}+\color{#00F}{16\sin^2\left(10^{\large\circ}\right)\sin\left(20^{\large\circ}\right)\sin\left(40^{\large\circ}\right)}\right)\tag{3c}\\[6pt]
&=AB^2\left(\color{#C00}{2-2\cos\left(20^{\large\circ}\right)}+\color{#090}{4\cos^2\left(20^{\large\circ}\right)-4\cos\left(20^{\large\circ}\right)+1}\right.\\
&\phantom{=AB^2\left(\right.}\left.\color{#00F}{{}-4\cos^2\left(20^{\large\circ}\right)+6\cos\left(20^{\large\circ}\right)-2}\right)\tag{3d}\\[12pt]
&=AB^2\tag{3e}
\end{align}
$$
Explanation:
$\text{(1a)}$: Law of Sines
$\text{(1b)}$: $\sin\left(150^{\large\circ}\right)=\frac12$
$\text{(2a)}$: Law of Sines
$\text{(2b)}$: Law of Sines
$\text{(2c)}$: $\sin\left(30^{\large\circ}\right)=\sin\left(150^{\large\circ}\right)=\frac12$
$\text{(3a)}$: Law of Cosines
$\text{(3b)}$: apply $\text{(1b)}$ and $\text{(2c)}$
$\text{(3c)}$: $\cos\left(100^{\large\circ}\right)=-\sin\left(10^{\large\circ}\right)$
$\text{(3d)}$: use $\sin(a)\sin(b)=\frac{\cos(a-b)-\cos(a+b)}2$ to get
$\phantom{\text{(3d):}}$ $\sin^2\left(10^{\large\circ}\right)=\frac{1-\cos\left(20^{\large\circ}\right)}2$ and
$\phantom{\text{(3d):}}$ $\sin\left(20^{\large\circ}\right)\sin\left(40^{\large\circ}\right)=\frac{\cos\left(20^{\large\circ}\right)-\frac12}2$
$\text{(3e)}$: simplify
