Can you please help me finding the value of angle $x$ in this image (I've drawn using microsoft paint, and added as many angles as many I could figure out). All angles are in degrees. Any exterior angle property or angle sum property seems not to help further.
Draw equilateral triangle $ACE$ such that $E$ lies on the same side of $AC$ as $B$. Then angle chasing shows that $\angle EAB = 18^\circ = \angle BAD$. Since $AC=BC=EC$, we have that $\angle ABE =\frac 12 \angle ACE = 30^\circ = \angle DBA$. Hence triangles $ABE, ABD$ are congruent by ASA. Therefore $AE=AD$, but $AE=AC$, so $AD=AC$. From this we get $\angle ACD = 90^\circ - \frac 12 \angle DAC = 78^\circ$.
Sorry for not marking the angles on the figure.