Some geometry

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Source: St Petersburg Olympiad 2012, Grade 9, P3

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#1 h Sep 29, 2017, 12:45 PM • 1 Y

Y by Adventure10

$ABCD$ is inscribed. Bisector of angle between diagonals intersect $AB$ anc $CD$ at $X$ and $Y$ . $M,N$ are midpoints of $AD,BC$ . $XM=YM$ Prove, that $XN=YN$ .

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#2 h Sep 10, 2023, 8:43 AM • 1 Y

Y by SolveForChocolate

In case $AB\parallel CD$ , symmetry wrt perpendicular bisector of $AB$ solves the problem. Now we will assume $AB$ and $CD$ are not parallel. Let $Q=AB\cap CD$ . By simple angle chasing, $QX=QY$ or $Q$ lies on the perpendicular bisector of $XY$ . Since $MX=MY$ , $M$ also lies on this perpendicular bisector. Note that $BC$ and $AD$ are antiparallel wrt $\angle AQD$ . This means that the line $QM$ , which is the median of $\triangle QAB$ is the symmedian of triangle $QBC$ . In other words, lines $QM$ and $QN$ are symmetric wrt bisector of line $AQD$ . If we look at it at another angle, $QM$ is the angle bisector of $\angle XQY$ (due to symmetry wrt perpendicular bisector of $XY$ ). So reflection of line $QM$ wrt angle bisector $\angle XQY$ should be $QM$ itself. But earlier, we found it is $QN$ . Thus, line $QM$ and $QN$ coincide or $Q,M,N$ are collinear. Since $Q$ and $M$ lie on the perpendicular bisector of $XY$ , $N$ also does. So $NX=NY$

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