Let $ABCDE$ be a cyclic pentagon, where $AC=2, AD=3, BD=5, BE=1, \frac{CD}{DE} = \frac{10}{3}$. What is the value of $\frac{BC}{CE}$?
I worked with the area of specific triangle with trigonometry. But I thought that it could arise a complex calculation. For my inability to draw the figure of the contextual problem, I was not sure whether it can give birth of paradox or not. But I am pretty sure there can be another way for solving this problem. In which way, should I go?
Please, pardon my error.