Geometry question involving triangles given with picture. Here's the question:


  
*
  
*$\overset{\Delta}{ABC}$ is a triangle.
  
*$D$ is a point on $[BC]$.
  
*$|BD|=4$.
  
*$|AD|=|CD|$.
  
*$\text m(\widehat{CBA})=\alpha=30^\circ$.
  
*$\text m(\widehat{ACB})=\beta=20^\circ$.
  
*What is $\color{magenta}{|AC|}=\color{magenta}x$?
  

Geometric approaches gave me nothing. Letting $|AD|=|CD|=a$, $|AB|=b$ and applying law of cosines to all three triangles yields;
$$
\begin{align}
x&=2a\cos 20^\circ\\
16&=a^2+b^2+2ab\cos 70^\circ\\
(a+4)^2&=b^2+x^2+2bx\cos 50^\circ
\end{align}
$$
respectively, from $\overset{\Delta}{ACD}$, $\overset{\Delta}{ABD}$ and $\overset{\Delta}{ABC}$. Directly isolating $x$ from these equations gives an expression involves trigonometric function(s), and i cannot find a way to annihilate trigonometric functions in these equations to reach the answer which is $\color{magenta}x=\color{magenta}4$.
 A: Here is an elegant purely geometric solution. The diagram shows the 18-gon that I used to find the solution, but the solution itself does not need the 18-gon.

Let $E$ be the image of $B$ under the reflection that maps $A$ to $D$.
Then $\angle EAC = \angle EAD + \angle DAC = \angle ADB + \angle DAC = 40^\circ + 20^\circ = 60^\circ$.
Also $\angle DEA = \angle DBA = 30^\circ$.
Thus $DE$ bisects $AC$ because $DE \perp AC$ and $\triangle ACD$ is isosceles.
Thus $\triangle EAC$ is equilateral and hence $\overline{AC} = \overline{AE} = \overline{BD}$.
A: First calculate some angles:
$ \angle DAC = \angle  DCA = 20^o $ (isosceles triangle theorem https://en.wikipedia.org/wiki/Isosceles_triangle )
$ \angle CAD = 180^o -20^o -20^o = 140^o $ (triangle is $180^o$)
$ \angle CAB = 180^o -30^o -20^o = 130^o $ (triangle is $180^o$)
$ \angle BAD = \angle BAC- \angle DAC = 130^o-20^o = 110^o $
and the rest is all the law of sinus ( https://en.wikipedia.org/wiki/Law_of_sines ):
$ \frac{AD}{\sin (\angle ABD)} = \frac{BD} {\sin(\angle BAD)} $
$ \frac{AD}{\sin (\angle ACD)} = \frac{AC} {\sin(\angle ADC)} $
done :) 
A: Drop a perpendicular from $D$ to $M$ on $AC$. By congruence of the resultant $\bigtriangleup AMD$ and $\bigtriangleup MDC\,(*)$ we observe that $|AM| = x/2$ and so $|AD| = (x/2)\sec20^{\circ}$.  Now we note that $A\hat{D}B = 20\times2 = 40^{\circ}$  since the exterior angle of a triangle equals the sum of $2$ opposite interior angles. So by the angle sum of $\bigtriangleup ABD,\,B\hat{A}D = 180-30-40 = 110^{\circ}$. Then, applying the sine rule: $$\frac{AD}{\sin A\hat{B}D} = \frac{4}{\sin B\hat{A}D}$$ we have $$ \frac{(x/2)\sec20^{\circ}}{\sin30^{\circ}} = \frac{4}{\sin 110^{\circ}}$$ which by rearrangement, yields $|AC|$:$$x = 8\sin 30^{\circ} \cos20^{\circ} \mathrm{cosec}\,110^{\circ} = 4$$
$(*)$ The congruence can be seen from the facts that:


*

*$|AD| = |DC|$

*$|MD|$ is a common length

*$A\hat{M}D = D\hat{M}C$ since $DM\perp AC$


satisfying the $RHS$ test for right triangles.
A: 
Construct isoceles $\triangle BDF$ with common angle $20^\circ$ and isoceles $\triangle BDG$ with common angle $30^\circ$. Now $\angle BGF=\angle FGD=\angle DGA=60^\circ$, and $\angle GDF=\angle ADG=10^\circ$. So $\triangle GDF\cong \triangle GDA$ and so $\overline{DF}=\overline{DA}$. Since $\triangle ADC\sim\triangle DFB$, we have the desired result.
