Geometrical problem on semi-circles

Question

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.

Answer 1

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.

Answer 2

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?

Answer 3

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

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