self polar triangle with respect to a circle

Question

I was reading this document but was stuck at the first lemma..

Can someone explain why harmonic sequence led to that conclusion?

Alternative proof is welcome too.

Answer 1

By definition, point $X$ is on the the polar of $R$ if $Y,Z$ are the intersection points of $RX$ with the circle and $(R,X;Y,Z)$ is harmonic. Since the points $E$ and $F$ satisfy this condition, the line $EF$ is the polar of $R$.

The straight lines in the figure are the complete quadrangle for the points $A,B,C,D$. By the projective properties of the complete quadrangle you can make symmetric arguments for $P$ and $Q$.

Answer 2

$\textbf{Hint:}$Draw tangents from $R$ to the circle,say at $X,Y$.Then,$(R,XY \cap AD,D,A)$ form a harmonic bundle.(This is true due to properties of symmedian)

It is a property that,if A-symmedian of a triangle $ABC$ meets its circumcircle at $K$.Then, $\frac{AB}{BK}=\frac{AC}{CK}$.In other words,$ABCK$ is a harmonic quadrilateral.Then you can take perspectivity at $B$ to show the above fact.

But by the conclusion that $(R,E;A,D)$ harmonic we have that $E=XY\cap AD$.Hence it lies on the polar of $R$.Similarly $F$ also lies on the polar of $R$.From which the conclusion will follow.