AC // BM wanted, right triangle, projections, midpoints, circumcenter

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AC // BM wanted, right triangle, projections, midpoints, circumcenter

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Source: 2017 Grand Duchy of Lithuania, Mathematical Contest p3 (Baltic Way TST)

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#1 h Oct 3, 2020, 5:29 PM

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Let $ABC$ be a triangle with $\angle A = 90^o$ and let $D$ be an orthogonal projection of $A$ onto $BC$ . The midpoints of $AD$ and $AC$ are called $E$ and $F$ , respectively. Let $M$ be the circumcentre of $\vartriangle BEF$ . Prove that $AC\parallel BM$ .

This post has been edited 1 time. Last edited by parmenides51, Nov 17, 2022, 8:46 PM
Reason: latex

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#3 h Oct 3, 2020, 6:01 PM

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Complex numbers seem to work fine, but I present a normal solution.

Claim. $EDB \sim FAB$
Obviously, $ADB\sim CAB$ , thus $\frac{AB}{DB}=\frac{AC}{DA}=\frac{AF}{DE}$ and because $\angle EDB = \angle FAB$ , the claim follows. $\quad \square$

Thus, $\angle FBA = \angle EBD$ .
Also, $\angle BXF=180^{\circ} - \angle BEF=90^{\circ} - \angle BED =\angle EBD=\angle FBA$ ( $X$ is arbitrary point on opposite from $\overarc{BEF}$ ), therefore $AB$ is tangent to $(BEF)$ .
Hence, $AC \perp AB\perp BM\implies AC\parallel BM$ . We are done.

This post has been edited 1 time. Last edited by rafaello, Oct 3, 2020, 6:36 PM

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#4 h Apr 2, 2021, 8:23 PM

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It is clear that $\triangle ABC \sim \triangle DBA$ and therefore $\triangle CBF \sim \triangle ABE$ . And since $CD||EF$ we have
$\angle ABE=\angle CBF=\angle EFB$
So $AB$ is tangent with $(BEF)$ and we are done.

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