Find an angle of an isosceles triangle

I saw the following solution may years ago:

On side $AD$ construct in exterior equilateral triangle $ADE$. Connect $BE$.

Then $AB=AC, AE=BC, \angle BAE=\angle ABC$ gives $\Delta BAE =\Delta ABC$ and hence $AB=BE$.

But then $$AB=BE, BD=BD, DA=DE \Rightarrow ADB =EDB$$

Hence $\angle ADB=\angle EDB$. Since the two angles add to $300^\circ$ they are each $150^\circ$. Then $\angle ABD + \angle ADB+ \angle BAD=180^\circ$ gives $ABD=10^\circ$.

One way to calculate this is to write sin laws for two triangles $ABD$ and $BDC$. Call the angle $\angle ABD=x$. Then we have: $$ \frac{AD}{\sin x}=\frac{BD}{\sin 20},\frac{BC}{\sin (20+x)}=\frac{BD}{\sin 80} $$ Using $AD=BC$ and $\sin 80=\cos 10$ we get the following: $$ \frac{\sin x}{\sin 20}=\frac{\sin (20+x)}{\sin 80}\implies \sin x=2{\sin 10}\sin (20+x)\implies\\ \tan x=\frac{2\sin 10\sin 20}{1-2\sin 10\cos 20} $$ Now consider the following identities: $$ 1-2\sin 10\cos 20=1-(\sin 30-\sin 10)=\frac{1}{2}+\sin 10=2\cos 10 \sin 20 $$ Replacing this result in previous equation we get: $$ \tan x=\frac{2\sin 10\sin 20}{2\cos 10\sin 20}=\tan 10 \implies x=10 $$ and hence $\angle DBC=70$.