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BAYMAX is cyclic!
Posted: Mon Apr 09, 2018 5:06 pm
Let $M$ be the midpoint of the side $AC$ of triangle $ABC$. Let $P$ on $AM$ and $Q$ on $CM$ be such that $PQ=AC/2$. Let $(ABQ)$ intersect with $BC$ at $A_X$ other than $B$. $(BCP)$ intersects with $BA$ at $A_Y$ other than $B$. Prove that $BA_YMA_X$ is cyclic
Re: BAYMAX is cyclic!
Posted: Mon Apr 09, 2018 5:21 pm
Let $N$ and $R$ be the midpoints of $AB$ and $BC$ resp.
Then we get $NRQP$ is a parallelogram.
Again, $AA_Y.AB=AP.AC$ means, $AA_Y.AN=AP.AQ$
It gives $NA_YPQ$ cyclic.
Similarly, $QMRA_X$ is cyclic.
We are working with directed angle
So, $BA_YMA_X$ is cyclic