Given the inscribed quadrilateral prove that $\frac{ab}{cd}=\frac{y}{x}$
I can get quite a few similar triangles due to several angles inscribing the same arc and the vertical angle. I also have by the power of the point that $(y)(x)=(EC)(AE)$. Now I'm going in circles with the similar triangles so is there another approach I'm not quite seeing in order to get the relationship I want?
We have two pairs of similar triangles (use angle-angle-angle): \begin{align*} \Delta AEB \sim \Delta DEC\\[4pt] \Delta BEC \sim \Delta AED\\[4pt] \end{align*} Thus, letting $v=CE$, we get \begin{align*} \frac{a}{c}&=\frac{y}{v}\\[4pt] \frac{b}{d}&=\frac{v}{x}\\[4pt] \end{align*} hence $$ \frac{ab}{cd} = \left(\frac{a}{c}\right)\!\left(\frac{b}{d}\right) = \left(\frac{y}{v}\right)\!\left(\frac{v}{x}\right) = \frac{y}{x} $$