Suppose $\triangle ABC$ is acute, and $\angle A = 30^\circ.$ Then, draw a circle with $BC$ as the diameter. Let the intersection points of the circle drawn and $AB$ and $AC$ be $D$ and $E,$ respectively. Find $S_{ADE}:S_{DBCE}.$
I first made a diagram, and I drew out the points. However, I wasn't quite sure where to go from here, although I was thinking of trying to use the fact that $\angle A = 30^\circ.$ Can I have a hint please?
Since $\measuredangle ADE=\measuredangle ACB, $ we see that $\Delta ABC\sim\Delta AED.$
Also, $\measuredangle BED=90^{\circ}.$
Thus, $$\frac{S_{\Delta ADE}}{S_{BDEC}}=\frac{S_{\Delta ADE}}{S_{\Delta ABC}-S_{\Delta ADE}}=\frac{1}{\frac{S_{\Delta ABC}}{S_{\Delta ADE}}-1}=\frac{1}{\left(\frac{AB}{AE}\right)^2-1}=\frac{1}{\left(\frac{2}{\sqrt3}\right)^2-1}=3$$