Beautiful cyclic quadrilateral property involving isogonal conjugates.

Question

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!

Answer 1

Let $DD \cap BB=H$,$AD \cap BC=F$,$AB \cap CD=E$. It is trivial that these points are collinear by pascal on $(ABB CDD)$.

now we only need to prove that $\triangle AP'D$ and $\triangle BCQ'$ are perspective so by Desargues theorem we have to prove that $DP' \cap BQ'$, $F$ and $H$ are collinear. we prove that $G,H,F$ and $G$ lie on a line. let $BQ'\cap \overline{FHE}=G$ and $DP' \cap \overline{FHE}=G'$ note that if we move $P$ with the same velocity, then $P \mapsto P' \mapsto G$ and $P \mapsto Q \mapsto Q' \mapsto G'$ so indeed we have $G \mapsto G'$ we only have to verify problem for 3 values of $P$.

• $P$ on $M$ then $G$ and $G'$ both become on $F$
• $P$ on $A$ then $G$ and $G'$ both become on $BD \cap FH$
• $P$ on the reflection of $A$ through $M$ and $Q$ on the reflection of $B$ through $M$ then both $G$ and $G'$ become on $H$.

so we are done. diagram