In Fig : what is $\angle DAC= ?$ In Fig : what is $$\angle DAC= ? $$ 

in triangle $\triangle ABC $ And $|AB|=|AC|$And Point $D$ in the triangle $\triangle ABC $ 
such that : $\angle BCD =40,\angle CBD =20$
My Try:
$$\frac{\sin 30}{AD}=\frac{\sin \angle BAD}{BD}$$
and :
$$\frac{\sin 10}{AD}=\frac{\sin \angle DAC}{DC}$$
So :
$$\sin \angle DAC =\frac{\sin 10}{\frac{\sin 30 \times BD}{\sin \angle BAD}}\times DC$$
$$\sin \angle DAC =\frac{\sin 10 \times \sin \angle BAD}{\sin 30 \times BD}\times DC$$
Now what?
 A: 
$E$  is the intersection between the line through $CD$ and $AB$.
It means that $∠BDE=60°→∠BED=90°$. Let's call $AB=AC=l$ and $BD=2p$.
Considering the right triangle $△AEC$ we have $AE=l\sin10°$.
Considering the right triangle $△BED$ we have that $BE=p\sqrt{3}$ and $ED=p$. Then
$$BE=p\sqrt{3}=AB−AE=l−l\sin10°→p=\frac{l(1−\sin10°)}{\sqrt{3}}$$
$$\tan(∠EAD)=\tan(80°−x)=\frac{ED}{AE}=\frac{1−\sin10°}{\sqrt{3}\sin10°}≈2.7475=\tan(70°)$$
I used Wolfram for the last equality, so $80°−x=70°→x=10°80°−x=70°→x=10°$.
A: Let $x = \angle DAB, y = \angle DAC$. Since sum of all angles of the outer triangle is $360^\circ$, so $x + y = 80$. You can get another equation using the sine rule for the sides $BD, AD, CD$;
$$ 1 = \frac{BD}{AD} . \frac{AD}{CD} . \frac{CD}{BD} = \frac{\sin x}{\sin 30} . \frac{\sin 10}{\sin y} . \frac{\sin 20}{\sin 40}; $$
so solving the two equations $$\frac{\sin x}{\sin y} = \frac{\sin 30 \sin 40}{\sin 10 \sin 20}$$ and $$x+y = 80$$ will give you the angle $y$. 
A: Let $E$ be a point on $BC$ such that $CD=DE$. Then $\angle CED=\angle DCE=40^\circ$ so $\angle BDE = \angle CED-\angle EBD=40^\circ-20^\circ=20^\circ=\angle EBD$ and therefore $EB=ED$.
Let $F$ lie on $BD$ and $DF=CD$. Since $\angle CDF = \angle DCB+\angle CBD =40^\circ+20^\circ=60^\circ$, triangle $CDF$ is equilateral. It follows that $CF=CD$. Easy to calculate $\angle FCA=50^\circ=\angle EBA$ and therefore triangles $EBA, FCA$ are congruent (SAS). Hence $\angle AFC =\angle AEB=180^\circ -\angle CEA$ which shows that points $A, F, C, E$ lie on a circle. The center of this circle is $D$ since $DE=DC=DF$. It follows that $DC=DA$ which leads to $\angle DAC=\angle ACD=10^\circ$.

