$ABCD$ is a cyclic quadrilateral. The midpoints of the diagonals $AC$ and $BD$ are respectively $P$ and $Q$. If $BD$ bisects $\angle AQC$, the prove that $AC$ will bisect $\angle BPD$.
Problem
Source: NMTC Junior 2023 P8
Tags: geometry, nmtc, cyclic quadrilateral
Ciel_vert
08.11.2023 16:06
Blatantly follows from the properties of harmonic quadrilaterals
SatisfiedMagma
24.11.2023 22:05
I think for a well-trained olympian, its very easy to see this is just converse of Humpty Point configuration, so projective does overkill. It took me personally quite some time to find it out, but I do admit that the problem becomes very trivial with projective technology. I will leave the synthetic solution for someone else to post. Solution: Consider the point $A'$ on the arc $ADB$ containing $A$ such that $AA' \parallel BD$ and let $AQ \cap \odot(ABCD) = C' \ne A$. The major claim is to basically show that quadrilateral $ABCD$ is also harmonic. Observe \[-1 = (D,B;Q, \infty) \overset{A}{=} (D,B;C',A').\]Since $AA'BD$ is a isosceles trapezium, by symmetry we would have $\angle CQD = \angle DQA = \angle A'QB$. This ultimately leads to the conclusion that $A'$ lies on the line $CQ$. By clear symmetry around the perpendicular bisector of $\overline{BD}$, we can conclude that \[-1 = (D,B;C',A') = (D,B;C,A)\]proving the main claim. It is a well-known property of Humpty Points now that $PA$ is an angle bisector of $\angle DPB$ and we're done. $\blacksquare$