In the inscribed quadrilateral $ABCD$, $P$ is the intersection point of diagonals and $M$ is the midpoint of arc $AB$. Prove that line $MP$ passes through the midpoint of segment $CD$, if and only if lines $AB, CD$ are parallel.
Problem
Source: Lotfi Zadeh Olympiad 2021, Problem 1
Tags: geometry, Parallel Lines, cyclic quadrilateral, Lotfi Zadeh MO
Jalil_Huseynov
28.12.2021 22:24
Solution that I found during contest: $\bullet$ If part: If $AB||CD$, then $MP$ is perpendicular bisector of $AB$, so it's perpendicular bisector of $CD$,too. So done. $\bullet$ Only if part: Assume $MP\cap CD=N$ and $CN=DN$. Let $\angle ABD=\angle ACD=\beta, \angle BAC=\angle BDC=\alpha, \angle CPN=\angle APM=\gamma, \angle DPN=\angle BPM=\theta$ and $\angle MAB=\angle MBA=\phi$. So Sine Law in $\triangle DPN$ and $\triangle CPN$ gives that $\frac{\sin \theta}{\sin \gamma}=\frac{\sin \alpha}{\sin \beta}$. And also Sine Law in $\triangle PAM$ and $\triangle PBM$ gives that $\frac{\sin \theta}{\sin \gamma}=\frac{\sin \beta +\phi}{\sin \alpha+\phi}$ $\implies$ $\frac{\sin \alpha}{\sin \beta}=\frac{\sin \beta +\phi}{\sin \alpha+\phi}$. So from here we can easily get that $\alpha=\beta$. So $AB||CD$.Done!
X.Allaberdiyev
05.08.2023 10:19
Part a) If $AB//CD$ then $MP$ is perp. bisector of $AB$, then it is also perp. bisector of $CD$, as desired. Part b) Assume that $MP$ passes through midpoint of $CD$, $BD$ and $CM$ intersect at point $K$ and, $MD$ and $AC$ intersect at point $L$, then from cheva theorem in triangle $MCD$ we have $KL//CD$ and we know that $\angle MCA=\angle MDB$, then $KLDC$ is isosceles trapezoid, then $\angle BDC=\angle ACD$, implies that $ABCD$ is isosceles trapezoid too, which completes the proof.