Consider a quadrilateral $ABCD$ in which the properties listed in the figure below are given.
We want to calculate the length of $BC.$
Using the first and two of the three angle properties it's easy to see that $BCD$ is isosceles, since
$$\angle{CDB}=\pi-\angle{CBD}-\angle{BCD}=\pi-\beta-2\alpha=\beta$$
and hence $\overline{BC}=\overline{CD}.$ I really can't find anything else. Using Geogebra it seems like the figure "wants" to be cyclic and to be an isosceles trapezoid, but if I use the angle relations to build it I don't have the correct lengths of $AX$ and $CX.$ I tried to prove that $ABCD$ is cyclic, but I can't find anything useful, I tried to use $CX$ and $AX$ as cevians and tried to use Stewart's theorem to get some relations with the sides, but I didn't have too much success. Can I have a solution, or at least some hints? In all my reasoning I didn't use the identity $\gamma+2\delta=\pi/2:$ maybe it can be used with some reflection or something like that, but all the constructions I've made were useless…
Best Answer
Let $M$ be the symmetric of $A$ with respect to $BD$. $M$ is on the circle with center $C$ and radius $R=BC$ since angle $BMD =$ half the central angle. Angle $BDM = \delta$ by symmetry, therefore angle $XCM=XCB+BCM=\gamma + 2 \delta $ is a right angle.
$R^2=65^2-33^2$
$R=56$