# Why are these angles equal?

-see picture above- (from Billiards and Geometry by Serge Tabachnikov)

I don't understand why the angles $$F_2BA_1$$ and $$F_1BA_0$$ are equal (I do understand the conclusion, that follows from the optical property of the ellipse). Can somebody help me out?

• Please don't vandalize your own question. – YuiTo Cheng May 26 at 9:06

$$\angle F_2BA_1=\angle F_1'BA_0$$ (vertical angles);
$$\angle F_1'BA_0=\angle F_1BA_0$$ (corresponding angles in the reflection about $$A_0A_1$$).
Hence, by transitivity, $$\angle F_2BA_1=\angle F_1BA_0$$.