enter image description here

-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?

  • 1
    $\begingroup$ Please don't vandalize your own question. $\endgroup$ – 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, $\angle F_2BA_1=\angle F_1BA_0$.


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