TITLE: Geometrical problem on semi-circles
QUESTION [1 upvotes]: 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.

REPLY [1 votes]: As suggested in the comments, I would start considering the center of the segment $[CB]$ (denoted by $P$). Since $\overset{\frown}{CD}:\overset{\frown}{DB}=3:10$, the first observation is that $$\angle DPC=\frac{180°·3}{13}=\frac{540°}{13}\quad\quad \angle BPD=\frac{180°·10}{13}=\frac{1800°}{13}$$
Therefore $$\angle PCD=\frac{180°-\angle DPC}{2}=\frac{900°}{13}\quad \quad \angle DBP=\frac{180°-\angle BPD}{2}=\frac{270°}{13}$$

Now consider the following proposition which I first learned in Evan Chen's exceptional 'Euclidean Geometry in Mathematical Olympiads' (EGMO).

Proving $(i)\Rightarrow (iii)$ is quite simple.

 Hint: Move point $C$ in the circumcircle so that $\angle BAC=90°$

What angles can now be determined?
We can now start with the conclusions. Observe that $$\angle CAD=\angle PCD-\angle ADC=\frac{900°}{13}-\frac{270°}{13}=\frac{630°}{13}$$
Thus, if you consider the segment $[OE]$, you'll obtain the following $$\angle EPA=180°-2·\angle CAD=\frac{1080°}{13}\quad\quad \angle BPE=180°-\angle EPA=\frac{1260°}{13}$$ Hence

$$\frac{\overset{\frown}{AE}}{\overset{\frown}{EB}}=\frac{\angle EPA}{\angle BPE}=\frac{1080°}{1260°}=\frac{6}{7}$$
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