# School geometry: Golfing exercise

I have the following exercise to solve:

A (golf) player hits the ball at B, the ball touches line a exactly once and goes into the whole L. How does the ball's way look like?

My idea has been the following (and I have forgotten most of my basic school geometry): The incoming angle of the ball must be equal to the outgoing one, hence I was just dividing the distance from B to L to determine where the ball must hit a. But somehow it feels that I am missing something, but I don't know what...any help appreciated.

## 1 Answer

HINT:

draw $L'$, reflection of $L$ across $a$, and join $B$ with $L'$.

To see why that works, see the diagram below: right triangles $ALH$ and $AL'H$ are congruent (SAS), whence $\angle LAH \cong \angle L'AH$. On the other hand we have $\angle L'AH \cong \angle BAa$ for they are vertical angles. It follows that $\angle LAH \cong \angle BAa$, as required.

• Many thanks. I have seen this solution, but why is my approach wrong, and also I would like to understand why your solution is the suggested one - what are the underlying rules and laws applied? I mean I assume that connecting B with L' will ensure that the incoming and outgoing angle of the ball hiting a are equal, but from what does that follow? Many thanks – Pugl Nov 20 '16 at 9:17
• I'm only suggesting a practical method to construct the bouncing point. To understand why it works only a little geometry is needed: mainly vertical angles and congruence criteria. I can provide a full explanation if needed. – Aretino Nov 20 '16 at 11:06
• I would really appreciate that! – Pugl Nov 20 '16 at 14:50
• I updated my answer. – Aretino Nov 20 '16 at 16:39
• ..now I don't even know what I did not get:) Maaany thanks! – Pugl Nov 20 '16 at 17:21