Triangle $ABC$ has circumcenter $O$. $M$ is the midpoint of arc $BC$ not containing $A$. $S$ is a point on $(ABC)$ such that $AS$ and $BC$ intersect on the line passing through $O$ and perpendicular to $AM$. $D$ is a point such that $ABDC$ is a parallelogram. Prove that $D$ lies on the line $SM$.
Problem
Source: IMOC 2023 G1
Tags: geometry, parallelogram
trinhquockhanh
09.09.2023 15:57
CrazyInMath
09.10.2023 20:50
Let $AS\cap BC$ be $T$, midpoint of $BC$ be $N$ and $AM\cap BC$ be $Q$. Clearly we have $OPQN$ cyclic
As $\measuredangle QAO=\measuredangle QTO$, $QOAT$ is cyclic.
Now we know that $\measuredangle SMA=\measuredangle SAM+\measuredangle MSA$
and $\measuredangle SAM=\measuredangle TAQ=\measuredangle TOQ=\measuredangle POQ=\measuredangle PNQ$
and $\measuredangle MSA=\measuredangle MBA=\measuredangle MBC+\measuredangle CBA=\measuredangle MAC+\measuredangle CBA=\measuredangle BAM+\measuredangle CBA=\measuredangle CQA=\measuredangle NQP$
So $\measuredangle SMA=\measuredangle NPQ$, so $SM\parallel NQ$ and we're done.
After reading the official solution, I found that there is an extremely easy way to finish with simson line, and I used messy angle chasing...