Cyclic quadrilateral given. Parallelity wanted

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Cyclic quadrilateral given. Parallelity wanted

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Source: Lotfi Zadeh Olympiad 2021, Problem 1

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#1 h Dec 28, 2021, 7:05 PM • 2 Y

Y by Jalil_Huseynov, parmenides51

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.

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#2 h Dec 28, 2021, 7:24 PM • 1 Y

Y by Tellocan

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!

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#3 h Aug 5, 2023, 7:19 AM

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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.

This post has been edited 1 time. Last edited by X.Allaberdiyev, Aug 17, 2023, 4:51 PM

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