Let $P \left (ct, \frac{c}{t} \right )$ be on the hyperbola $xy=c^2$. Points $F$ and $G$ are the $x$ and $y$ intercepts of the tangent drawn from $P$, respectively. From $P$, a normal is drawn to intersect $y=x$ at $H$.
Prove that $FH \perp GH$.
Question:
How can we prove this without bashing out analytic geometry? I am attempting to prove this using purely geometric methods.
My Idea:
It suffices to show that $OFHG$ is cyclic. There are a few ways I can see of doing this, but my attempts have not been successful.
We know that $\measuredangle HOF = \frac{\pi}{4}$ so to complete the proof, we need only to show something like $\measuredangle HGF = \frac{\pi}{4}$ or a similar result.
$P$ is the midpoint of $GF$ (standard result of the rectangular hyperbola), so we need only to show that $HP$ is also equal to the radius of the circumcircle of $\triangle GOF$.
From $FP=GP$ and $FG \perp HP$ we conclude that $HP$ is the perpendicular bisector of $FG$. Therefore $FH=GH$.
Let $H'$ be the midpoint of arc $FG$ (not containing $O$) of the circumcircle of $FGO$. On the one hand, $\angle FOH'=\angle H'OG$ because arcs $FH'$, $GH'$ are equal. It follows that $H'$ lies on the bisector of angle $FOG$, i.e. on the line $HO$.
On the other hand, since arcs $FH', GH'$ are equal, we have $FH'=GH'$, therefore $H'$ lies on the perpendicular bisector of $FG$, i.e. on the line $PH$.
Thus $H' \in HO \cap HP = \{H\}$, i.e. $H=H'$. At this point we know that $H$ lies on the circle with diameter $FG$. Therefore $FH \perp GH$.