If a quadrilateral $ABCD$ is harmonic (which implies it is cyclic by
definition), then $AC$ is a symmedian in triangles $BAD$ and $BCD$ and
$BD$ is a symmedian in triangles $ABC$ and $ADC$.
This statement can be found on page 151 in the book Lemmas in Olympiad Geometry by Titu Andreescu. I can't prove it and there is no proof in the book. My first idea was to let $\{O\}=AC\cap BD$ and then try to prove that $$\frac{AO}{OC}=\frac{AB^2}{BC^2}$$
which implies $BD$ is a symmedian in triangle $ABC$. The other symmedians can be proven similarily. I don't know how to prove what I want to prove. Can you please help me? Thank you in advance!
Best Answer
Let the $A$ symmedian of $\triangle ABD$ intersect $\odot (ABCD)$ at $C'$. Then $(AC',BD)=(AC,BD)=-1$ and thus $C'\equiv C$. Do similarly for other triangles.