ABCD is a cyclic quadrilateral with AB = AD and ∠ BAD = 108°. Find ∠ BCD, ∠ ABD and ∠ ACB.

Step 1. The angle BCD. $\angle BCD = 180^\circ - 108^\circ = 72^\circ$.

Step 2. The angle ABD. In the triangle $ABD$, $AB = AD$, so $\angle ABD = \angle ADB$. They add up to $180^\circ - 108^\circ = 72^\circ$. So each is $36^\circ$.

Step 3. The angle ACB. It stands on the chord $AB$, like $\angle ADB$. So $\angle ACB = 36^\circ$.

Step 4. Notice. $\angle ACD$ is $36^\circ$ too, so $CA$ cuts $\angle BCD$ in half.