Let $ABCD$ be a convex cyclic quadrilateral, such as triangles $BCD$ and $CDA$ are not equilateral. Prove that, if $A$-Simson line is perpendicular to Euler line in $BCD$, $B$-Simson line is perpendicular to Euler line in $ACD$.
I know the following definitions, but I am not so sure of converting them into complex / barycentric form:
- Simson line is defined for each point $P$ on the circumcircle of the triangle $ABC$. The projections of $P$ on triangle's sides are collinear on $P$-Simson line;
- Euler line is the line that contains $O, G, H, O_9$ for each triangle, with these points being in order circumcenter, centroid, orthocenter and the center of the Euler circle, containing the altitudes' feet, midpoints of triangle's edges and midpoints of segments between verices and $H$.
Thanks in advance!