Find the angles of an inscribed trapezoid (in a circle) $ABCD$ ($AB||CD$) if $\angle ABD = 63^\circ$.
Any trapezium in a circle is an isosceles trapezium, so $AD = BC$, thus $\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{AD} = \newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{BC} = 2\cdot63^\circ = 126^\circ$. I've tried to calculate some angles if $P = AC$ $\cap$ $BD$ : $\angle APD = 126^\circ$ and $\angle APB = 54^\circ$. It seems useless and I think that there's a missing information. Is is possible to solve the problem?
It is not possible to solve the problem with the given information. Any isosceles trapezoid can be inscribed in a circle. Then let's start with some given $AB$ segment, and we draw a line from $A$ and one from $B$ at the given angle, that will intersect at point $P$ in your figure. Now choose any point $D$ on the extension of $BP$, away from $B$, on the same side as $P$, then draw a parallel to $AB$. This will intersect the extension of $AP$ in $C$. You can immediately see that this is an isosceles trapezoid, that can be inscribed in a circle. Now choose a point $D'$, find $C'$, similarly to the procedure above. Once again $ABC'D'$ is an isosceles trapezoid, which can be inscribed in a circle, but $\angle BAD\ne\angle BAD'$. Therefore you cannot solve the original problem.