Sine Law or No?

Question

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?

Answer 1

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 $.

Answer 2

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, $\angle BDC = \angle KBM = 30^\circ$.