Find the sum of the angles The question is below.

The solution says:
Solution

My questions:
I'm wondering why it was concluded that the path goes around the center three times. Also, what does that have to do with the external angles?
 A: Draw a ray from anywhere in the open center region passing through the point $A$ and ending well outside everything in the figure. Then, starting at $A$, begin tracing the figure until you return to $A$ again. Count how many times you hit the ray you drew (don't count the starting point as a hit, but you do count the ending point as a hit). You will find you hit the ray three times -- once for every circuit around the central point you chose.
The external angles are how much you turn as you "round the corner" around the outside. If you go around once, you have turned $360^{\circ}$.
A: At Wikimedia there's a good animation of winding number. For external angles there's a picture of a figure where external angles add up to $360^\circ$. Similar thing can be seen when the path goes around more than once, in your example it goes around three times, so external angles add up to $3\cdot360^\circ$.

A: Envision a car that starts at $A$ driving towards $F$, then keeps going until it ends up back at $A$ again.
I suspect the idea that the path goes around the center three times is meant to be intuitive, though whether it is or not is up to debate. Regardless, it can be helpful to have a more rigorous way to tell these things. Here's one: consider some line passing through $A$ and going off somewhere between $F$ and $E$. Trace the path from $A$ with your finger, counting every time you cross it from right to left.
You'll get two, and since you start and end at $A$, count that as another crossing. So three.
Take a second to convince yourself that when the car turns around an angle, it rotates a number of degrees equal to the external angle to (which measures $180^{\circ}$ minus) that angle. 
For example, (not thinking about the above diagram at all), if the car is on Main St. and needs to take Elm St, which branches off slightly to the right,
   Elm
  /
 /
 |
 |
Main

then Elm St. makes a very wide angle with Main St., but the car itself doesn't need to turn very much.
But if the car needs to turn onto Crazy St., as below:
 | \
 |  \
Main \
 |    \
 |   Crazy

then the car needs to turn wildly, even though Crazy St. makes a very small angle with Main St.
If the car needs to make a U-turn, then it needs to turn a full $180^{\circ}$, since Main St. forms a $0^{\circ}$ angle with itself.
So for an internal angle $\alpha$ and external angle $\beta$ we have $180^{\circ} - \beta = \alpha$.
The amount the car turns is $3 \cdot 360^{\circ} = 1080^{\circ}$, because that's three full turns. That's also the sum of all the external angles. Summing everything up we get (because there are $9$ points):
$$
\begin{aligned}
9 \cdot 180^{\circ} - \text{sum of $\beta$'s} &= \text{sum of $\alpha$'s}\\
9 \cdot 180^{\circ} - 1080^{\circ} &= \text{sum of $\alpha$'s}\\
\end{aligned}
$$
Bonus: This idea can be used to show that the sum of the internal angles of a triangle is always $180^{\circ}$. Which already puts you ahead of the wisest geometers of the Dark Ages.
