I found this problem at an elementary school festival and I've spent the last 30min making no progress whatsoever. I have a suspicion that the solution involves cleverly adding a line to the diagram, but can't find anything useful. Could somebody kindly provide a hint as to how to proceed?
(The question is to find $x$ and $y$)
Let $ABCD$ be the quadrilateral where $A$ is the upper left point, $D$ is the upper right point.
$\triangle ABC$ is an isosceles triangle with $AB=BC$. Here, take a point $F$ on the line $DC$ such that $\angle{FBC}=20^\circ$.
Since $\angle{BCF}=\angle{BFC}=80^\circ$, we have $BC=BF$. Then, we know that $\triangle ABF$ is an equilateral triangle, so we have $BF=FA$.
Since $\angle{FBD}=\angle{FDB}=40^\circ,$ we have $BF=FD$.
Hence, since $FA=FD$, we know that $\triangle AFD$ is an isosceles triangle. So, we have $$\angle{AFD}=40^\circ,\ \ \angle{ADF}=70^\circ.$$ Hence, we have $$y=\angle{ADB}=70^\circ-40^\circ=30^\circ,\ \ \ x=\angle{CAD}=80^\circ.$$
