Geometrical problem on semi-circles Given that $AE$ is the tangent to the small semi-circle at $D$ and that arc $CD$ : arc $DB$ = $3 : 10$, find arc $AE$ : arc $EB$. 




How do I go about solving this? I do not know how to start.
connect BD, CD, BE, draw perpendicular line from D to AB as F, connect D with the centre of the smaller semi-circle O. from $\frac{CD}{BD}$, derive $\frac{CF}{BF}$. Together with CO=OB=OD, you can derive $\frac{OF}{OD}$ and $\frac{DF}{FO}. $ since OD is perpendicular to AE, BE is also perpendicular to AE, using similar triangle, $\frac{DF}{FO}$ =$\frac{AD}{DO}$=$\frac{AE}{BE}$, you get the answer.
Let $O$ be a center of the small circle.

Thus, $$\measuredangle COD:\measuredangle DOB=3:10$$ and since
$$\measuredangle COD+\measuredangle DOB=\pi,$$ we obtain:
$$\measuredangle COD=\frac{3\pi}{13},$$
$$\measuredangle EAB=\measuredangle DAO=\frac{\pi}{2}-\frac{3\pi}{13}=\frac{7\pi}{26}.$$
Can you end it now?