South-Afrika, 2011

Notice that $\triangle ABC \equiv \triangle BCD$ (s, s, s) and $\triangle ABD \equiv \triangle DCA$ (s, s, s).
Let $\alpha = \angle CBD = \angle BCA = \angle CAB = \angle CDB$
and $\beta = \angle ACD = \angle ADC = \angle BAD = \angle ABD$.

It can then be seen that $3\alpha + \beta = 180^\circ$ from triangle $ABC$ and $4\beta + 2\alpha = 360^\circ$ from quadrilateral $ABCD$. Solving these two simultaneous equations gives $\alpha = 36^\circ$ and $\beta = 72^\circ$. Hence $\angle A = 72^\circ$, $\angle B = 108^\circ$, $\angle C = 108^\circ$, $\angle D = 72^\circ$.