Let $P$ be a given point inside quadrilateral $ABCD$. Points $Q_1$ and $Q_2$ are located within $ABCD$ such that $\angle Q_1 BC = \angle ABP$, $\angle Q_1 CB = \angle DCP$, $\angle Q_2 AD = \angle BAP$, $\angle Q_2 DA = \angle CDP$. Prove that $\overline{Q_1 Q_2} \parallel \overline{AB}$ if and only if $\overline{Q_1 Q_2} \parallel \overline{CD}$
I saw its solution. It involves isogonal conjugates.I know that isogonal conjugates is the reflection of a point along each angular bisector,actually they coincide but I think this is not the case with a point outside the triangle.Here P' is the reflection of P across bisector of $\angle A$ but it is not the same point as reflection across other bisectors??
Solution states that points $P$ and $Q_1$ are isogonal conjugates with respect to triangle BCX.I don't think they are reflections.I don't understand how are they conjugates.Is there any flaw in my understanding?? Correct me?
Here is my picture of diagram
My comment is true.
Here's an example of the isogonal conjugates taken with respect to the internal angle bisector at A, and the external angle bisectors of B and C, for the point P. These 3 lines intersect at another point $P'$. (Apart from the external angle bisectors, I had to remove all the other lines because they were distracting)
Note: If $P$ was "too close" to $BC$, the $P'$ would be on the other side of $BC$.
I'm not fully certain what your first image is showing, because the construction steps are not clear. Can you elaborate on how you drew the lines $BP', CP'$?
In particular, $BP$ looks very close to the external angle bisector already, and my eyeballing of $BP'$ indicates that $BP$ and $BP'$ are not isogonal conjugates of the exterior angle bisector.
In fact, given how close your point $P$ is to $BC$, I would expect the conjugates to meet on the other side of $BC$.