Sine Law or No? The diagram shows a triangle $ABC$ where 
$$AB = AC,\, BC = AD \text{ and } \angle BAC = 20°.$$
Find $\angle ADB$.

I used the Sine Law; We know that $\sin(C)/\sin(BDC) = \sin(A)/\sin(ABD)$ If we let $\sin BDC = a$, then the equation will be equal to $\sin 80/\sin a = \sin20/\sin(a-20)$, I could not find any correlation with the the angle a and the other angles, is there a way to solve this using the sine law or is it a bad approach in general (or is the implementation of the sine law wrong?).  What other approach would be a lot more useful in this kind of problem?
 A: Ok, let start with your equation 
$$\frac{\sin 80}{\sin a}=\frac{\sin 20}{\sin(a-20)}. $$
Using twice equation $\sin(2x)=2\sin x\cos x $, we reach
$$\frac{4\sin 20\cos 20\cos 40 }{\sin a}=\frac{\sin 20}{\sin(a-20)} $$
or
$$\frac{4\cos 20\cos 40}{\sin a}=\frac{1}{\sin(a-20)}. $$
We know that $2\cos20\cos40=\cos60+\cos20 $, thus
$$\frac{1+2\cos 20}{\sin a}=\frac{1}{\sin(a-20)}. $$
Simplifying,
$$\sin a=\sin(a-20)+2\cos20\sin(a-20)=\sin(a-20)+\sin a+\sin(a-40) $$
which turns to
$$\sin(a-20)+\sin(a-40)=0. $$
This equation has $a=30$ as a solution, thus $\sphericalangle BDA=150 $.
A: You can solve it without any trig at all, but it requires some "inspired" additional constructions. Reflect the triangle across one of its sides, $AC$, and also let $BL$ be the angle bisector of $\angle ABC$:

All the labelled angles in the diagram are easy to calculate using properties of isosceles triangles and/or sums of angles in triangles, using the reflection symmetry and/or the bisector. For example, $\angle ABK = 70^\circ$ because $\Delta ABK$ is isosceles with vertex $\angle BAK = 40^\circ$. 
Now consider $\Delta KLM$ and $\Delta KLC$ - they are congruent by two angles and a side ($KL$ which they have in common), therefore $KM = KC = BC = AD$ (the blue segments; the second equality holds because of the reflection). Also, now we can see $BM = AM = CD$ (the first from the isosceles $\Delta ABM$; the second, from $AD + DC = AC = AB = AK = AM + MK$). 
Finally, see that $\Delta BCD$ and $\Delta KMB$ have two sides equal (the blue and red segments) and the $80^\circ$ angle between them as well, so they are congruent, too. From this, $\angle BDC = \angle KBM = 30^\circ$.
A: Alternative construction and demonstration with simple Euclidean geometry below. Sorry for the label change.

Here we have $AC \cong BC$, $CP \cong AB$ and $\angle ACB = 20°$.


*

*Draw two isosceles triangles congruent to $ABC$ and with one side in common, so that $\triangle ABC \cong \triangle BCD \cong \triangle DCE$. 

*Show that $\triangle ACE$ is equilateral.

*Determine $\angle EAB = \angle CAB - \angle CAE$.

*Demonstrate that $\triangle CPB \cong \triangle EAB$, with SAS criterion. Therefore $\angle CPB \cong \angle ABE$.

*Show that $\triangle DBE$ is isosceles.

*Thus calculate the measure of the angle $\angle DBE = \frac{180°-80°}{2}$.

*So finally $\angle CPB = \angle ABE = \angle ABD - \angle ABE = 150°$.



An even simpler construction is reported below ($\triangle ADB$ is equilater; show then that $\triangle ADC \cong \triangle CDB \cong \triangle CPB$ and deduce from there the result).

A: 
Construct equilateral triangle $ADE$ outside triangle $ABC$. Since $AE = AD = BC$ and $$\angle \, EAC = \angle \, EAD + \angle \, BAC = 60^{\circ} + 20^{\circ} = 80^{\circ} = \angle \, BCA$$ triangles $CEA$ and $ABC$ are congruent. However, $ABC$ is isosceles so $CEA$ is also isosceles where $CA = CE$ and $\angle \, ACE =\angle \, CAB = 20^{\circ}$. Moreover, by construction $AD = ED$ and $CD$ is a common side, so the two triangles $\Delta \, ACD$ and $\Delta \, ECD$ are congruent (in fact mirror symmetric with respect to the line $CD$). Hence $\angle\, ACD = \angle \,ECD = \frac{1}{2} \, \angle \, ACE = \frac{1}{2} \, 20^{\circ} = 10^{\circ}.$ Therefore   $$\angle \, ADC = 180^{\circ} - \angle \, CAD - \angle \, DCA = 180^{\circ} - 20^{\circ} - 10^{\circ} = 150^{\circ}$$ 
