One method of making a bank shot in billiards involves imagining two lines, the so called "cross-pocket" and "cross-ball" lines:
One then projects their point of intersection to the far rail:
The resulting "bank point" guarantees a successful trajectory with equal angles of incidence and reflection on the far rail. I'm trying to prove that these angles are in fact equal, without using coordinates. Here's a diagram of the scenario with relevant points labeled and line segments drawn:
Here's the problem: fix point O anywhere on the table, and construct all line segments as shown in the diagram above. Prove that ∠OPA = ∠CPB. Thanks for the help!
First two pictures are from a helpful article linked here.
From $$ BP:AB=IP:AO $$ and $$ AP:AB=IP:BC $$ one gets (dividing the first equality by the second one): $$ BP:AP=BC:AO. $$ Hence triangles $APO$ and $BPC$ are similar and have the same angles.