Recently I observed this interesting problem of an ellipse, but I could not prove it.
Let $\alpha$ be an ellipse, let $F$ be one of its focal points, with directrix $l$, let $A$ be the vertex on the major axis opposite to $F$. Choose an arbitrary point $P$ on $\alpha$, with tangent line $m$. If $AP \cap l = Q$ and $m \cap l = R$, then $RQ = RF$. (See following figure)
My attempt: If we connect $QF$ and draw its perpendicular line at $F$, then we obtain the polar line of $Q$, and its intersection with the tangent line of $A$ is the polar point $S$ of $AQ$. Then $P$ is the unique point on $AQ$ such that if $R$ is the intersection of $SP$ with $l$, then $RF \perp FP$. Therefore I reduced the problem to this:
Let $AS$ and $l$ be two lines perpendicular to $AF$, let $Q$ be a point on $l$ such that $FS \perp FQ$. Let $P$ be a point on $AQ$ and let the intersection of $SP$ and $l$ be $R$. Show that if $FP \perp FR$ then $RQ = RF$. However I could not prove this. Any ideas on how to progress? Thank you!
After many failed attempts, I discovered that the reduced problem could be solved via automated theorem proving program. I wrote one in Mathematica based on this paper: https://pdfs.semanticscholar.org/ff1f/50570f6d9cfde86f3220f37f65ae2ef7c64f.pdf