Let $ABCD$ be a quadrilateral inscribed in a circle $\omega$, and let $P$ be a point on the extension of $AC$ such that $PB$ and $PD$ are tangent to $\omega$. The tangent at $C$ intersects $PD$ at $Q$ and the line $AD$ at $R$. Let $E$ be the second point of intersection between $AQ$ and $\omega$. Prove that $B$, $E$, $R$ are collinear.
Re: APMO 2013 P5
Posted: Tue Nov 22, 2016 4:10 pm
by joydip
Let, $G=BD \cap AC,F=DC \cap AQ$ .Then $,(P,G;A,C)= -1 $.So , R,F,G are collinear.Applying pascals theorem on hexagon $AEBDCC$ we get $EB, CC$(tangent at $C),FG$ are concurrent $\Rightarrow B,E,R$ are collinear.
Take a projective transformation that fixes ω and sends the pointAC ∩ BD to the center of the circle. Thus ABCD is a rectangle. Because ABCD is harmonic, it must in fact be a square. Thus P is the point at infinity along AB CD and the problem is not very hard now. -Copied from Evan Chen's book