# Inscribed trapezoid in a circle

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.