# Beautiful cyclic quadrilateral property involving isogonal conjugates. Let $ABCD$ be a cyclic quadrialetral. Diagonals $AC$ and $BD$ intersect at point $S$.

Denote midpoint of $AC$ with $M$. Choose points $P\in MD$ and $Q\in MB$ so that $PQ\parallel BD$ (in other words $MP/MD=MQ/MB$).

Denote with $P'$ isogonal conjugate of $P$ with respect to triangle $ACD$ and with $Q'$ isogonal conjugate of point $Q$ with respect to triangle $ABC$.

Now prove that points $P'$, $S$ and $Q'$ are colinear.

My thoughts: I very much doubt that this problem has a nice and simple solution, because there is a lot of stuff to tackle. I have tried barycentric/trilinear coordinates, but it gets quite messy (in the end we need to check if one $3 \times 3$ determinant is zero). I would have also tried complex numbers, if I knew how to determine the position of isogonal conjugates. From non-computational techniques nothing comes to my mind.

I would appreciate if someone could solve this in a way that could be carried out by hand in a reasonable time (that's why I gave up on barycentric coordinates). Thanks!