Find an angle in a given triangle $\triangle ABC$ has sides $AC = BC$ and $\angle ACB = 96^\circ$. $D$ is a point in $\triangle ABC$ such that $\angle DAB = 18^\circ$ and $\angle DBA = 30^\circ$. What is the measure (in degrees) of $\angle ACD$?

Take $O$ the circumcenter of $\triangle ABD$, see that $\triangle DAO$ is equilateral and, since $\widehat{BAD}=18^\circ$, we get $\widehat{BAO}=42^\circ$, i.e. $O$ is reflection of $C$ about $AB$, that is, $AOBC$ is a rhombus, hence $AD=AO=AC,\triangle CAD$ is isosceles with $\widehat{ADC}=\widehat{ACD}=78^\circ$, done.
