Suppose I have a right triangle $\triangle BCD$, inscribed in a circle $Q$ with radius $s$, where $BD$ is the diameter of the circle. Solve for the lengths of the sides of $\triangle BCD$ in terms of $s$?
Can I answer the problem using only the information given?
Try: We know that $BD$ is $2s$ so: $$4s^2=BC^2+CD^2$$
Suppose $\angle DQC$ is $\theta$, I know that $\angle DBC=\frac\theta2$, $\angle\hat {QDC}=\angle\hat {QCD}=\frac{180-\theta}2, \ \angle BQC=180-\theta$ and $\angle QBC=\angle QCB = \frac\theta2$. I'm using the following equations, but I seem to be going around circles: $$\begin{align} \frac{\sin\theta}{CD}&=\frac{\cos\frac\theta2}{s}\\ BC^2&=2s^2-2s^2\cos\theta\\ \frac{\sin\frac\theta2}{s}&=\frac{\sin\theta}{BC} \end{align}$$