Argentine National Olympiad 2016

We have $\widehat{BAD}=180^{\circ }-(29^{\circ }+41^{\circ })=110^{\circ }$, $\widehat{BCD}=82^{\circ }+58^{\circ }=140^{\circ }$. Consider the circumcircle $\gamma$ of triangle $BCD$. Since $\widehat{BAD}+\widehat{BCD}>180^{\circ }$, point $A$ is interior to $\gamma$.

Extend $CA$ beyond $A$ to meet $\gamma$ at $E$. By inscribed angles $$\widehat{EBD}=\widehat{ECD}=\widehat{ACD}=58^{\circ },\widehat{EDB}=\widehat{ECB}=\widehat{ACB}=82^{\circ }$$ Given that $\widehat{ABD}=29^{\circ }$, $\widehat{ADB}=41^{\circ }$ we obtain that $BA$ and $DA$ are bisectors of $\widehat{EBD}$ and $\widehat{EDB}$ respectively. Hence $A$ is the incenter of triangle $BDE$, implying that $EA$ is the bisector of $\widehat{BED}$.

From the cyclic quadrilateral $BCDE$ we have $$\begin{array}{rl}\widehat{BED}& =180^{\circ }-\widehat{BCD}=180^{\circ }-140^{\circ }=40^{\circ }.\text{Therefore }\widehat{DBC}=\widehat{BEC}=\frac{1}{2}\widehat{BED}=20^{\circ }\text{ and analogously}\\ \widehat{DBC}& =20^{\circ }.\text{In conclusion }\widehat{ABC}=29^{\circ }+20^{\circ }=49^{\circ },\widehat{ADC}=41^{\circ }+20^{\circ }=61^{\circ }.\end{array}$$