Qusetion is from Intro to Analysis by Courant, s3.6-2. Both question and solution follows:
My issue is at the very end following "Note that...". How do we arrive at that equation with the sines? Anyone have alternative to proving the bisection of angle $FQP$?
Here's an alternative proof using basic calculus and Euclidean geometry. Since there's always variance in the notation for Euclidean geometry, $PQ$ will denote a segment from point $P$ to point $Q$ and $|PQ|$ will denote the length of that segment.
We know that the length of the straight line path is minimized by taking $Q(x,y=\eta)$, which means $PQ$ is parallel to $FN$. The slope of the normal line to the curve at $Q$ is given by $-\frac{dx}{dy}\big|_{(x,\eta)} = -\frac{\eta}{p}$, and so it's easy enough to compute that the intersection of the normal line and the $x$-axis is precisely $N(x+p,0)$. It is then straightforward computation using the distance formula to see that $|FQ|=|QN|$, whence $\triangle FQN$ is isosceles and so $\theta = \psi$. Since $\psi$ and $\phi$ are alternate interior angles, it follows that $\phi = \psi$ and therefore $\theta = \phi$ as required.