Synthetic geometric proof that intersections of three tangents of a hyperbola are at right angle with respect to the foci

Question

Here is the construction I am looking at.

Let $F_1$ and $F_2$ with respective vertices $V_1$ and $V_2$. Let $P$ be a point be closer to $F_2$ than $F_1$. Tangents are drawn from from $V_1,$ $V_2,$ and $P.$ As $V_1$ and $V_2$ are parallel the only intersections between the tangents come from the tangents $P/V_1$ and $P/V_2.$ Let these intersections be $Q$ and $R.$ Now, $\angle RF_1Q=90^{\circ}.$

Here is a picture:

This is the fact I want to prove geometrically. Most likely a straightforward analytical solution exists but I am more interested in how to approach this synthetically.

I've tried angle chasing based on the fact that the tangent $P$ bisects the angle $\angle F_1PF_2.$ But both my attempts at finding $\angle RF_1Q$ directly and trying to prove $F_1RF_2Q$ concyclic have failed.

Similarly, I've had trouble trying to find more about geometric properties of hyperbolas online as most websites discuss hyperbola using coordinates which I am trying to avoid here.

Answer 1

Before the main proof, we need a nice lemma:

The two tangents that can be drawn to a hyperbola from an external point subtend equal or supplementary angles at the focus, according as the points of contact are on the same or opposite branches of the curve.

I'll prove this lemma for the same branch case, the other case is similar. Let then $OQ$, $OQ'$ be two tangents from point $O$ to a hyperbola of focus $S$. Let $Z$ be the intersection point of line $OQ$ with the directrix: we have then $SZ\perp SQ$ (see here for a proof). Drop from $O$ the perpendiculars $OU$ to $SQ$ and $OI$ to the directrix (see figure below) and let $M$ be the projection of $Q$ on the directrix.

From similar triangles we have then:

$$ SU:SQ=ZO:ZQ=OI:QM, $$ that is: $$ SU={SQ\over QM}OI=e\,OI, $$ where $e$ is the eccentricity of the hyperbola. Considering tangent $OQ'$ we can analogously prove $SU'=e\,OI$, where $U'$ is the projection of $O$ on $SQ'$. Hence $SU=SU'$ and triangles $OSU$, $OSU'$ are congruent. It follows that $\angle OSQ=\angle OSQ'$, as it was to be proved.

Let's prove now the requested property $\angle QF_1R=90°$.

From the above lemma we get: $$ \angle V_1F_1R=\angle RF_1P, \quad\text{and}\quad \angle QF_1V_1+\angle QF_1P=180° $$ and we have then: $$ \begin{align} 2\angle QF_1V_1+\angle V_1F_1P &=180°\\ 2\angle QF_1V_1+2\angle V_1F_1R &=180°\\ \angle QF_1V_1+\angle V_1F_1R &=90°\\ \angle QF_1R &=90°\\ \end{align} $$ QED.