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Midpoint Circle Dance


In an acute triangle $ \triangle ABC $, the midpoint of $ BC $ is $ M $. Perpendicular lines $ BE $ and $ CF $ are drawn respectively on $ AC $ from $ B $ and on $ AB $ from $ C $ such that $ E $ and $ F $ lie on $ AC $ and $ AB $ respectively. The midpoint of $ EF $ is $ N $. $ MN $ intersects $ AB $ at $ K $. Prove that the four points $ B, K, E, M $ lie on the same circle. 


Source: BdMO 2022 National Secondary P5


Proof Based Problems  


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Solution

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$\angle BEC = \angle CFB = 90$

So, $BCEF$ is cyclic. Let this circle be $\Gamma$

$BC$ is the diameter of $\Gamma$ because $BEC = 90$

So, $M$ is the center of $\Gamma$

As $N$ is the midpoint of arc $EF$, $MN$ is perpendicular bisector of $EF$.

Now, $\angle EMK = \angle EMN = \frac{1}{2}\angle FME = \angle FBE = \angle KBE$

So, we have $\angle EMK = \angle EBK \implies EBKM$ cyclic.


(Fun fact: $A$ was not used in this solution.)

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