# (Euler line) Prove that the orthocenter M, the center O of the circumscribed circle and the barycenter S are collinear. The point S divides the segment OM in the ratio 1:2.

(Euler line) Prove that the orthocenter M, the center O of the circumscribed circle and the barycenter S are collinear. The point S divides the segment OM in the ratio 1:2.
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Cristiano Sears

Step 1
We know that, the line on which orthocenter, circumcenter and barycenter(centroid) lie is called Euler Line of the triangle.
Now let a triangle $\mathrm{△}ABC$ and its medial triangle is $\mathrm{△}DEF$

Here O is orthocenter, M is circumcenter and S is barycenter.
Here we can say $\mathrm{△}ABC$ is similar to $\mathrm{△}DEF$ because D, E and F are the midpoint of lines BC, AC and AB respectively.
And also
$BC\mid \mid EF$
$AC\mid \mid FD$
$AB\mid \mid ED$
Hence
$\mathrm{△}ABC\sim \mathrm{△}DEF$ with 2:1 ratio.
In the above figure O is the circumcenter of $\mathrm{△}ABC$ which is also the orthocenter of $\mathrm{△}DEF$.
Now we have to prove O, S and M are collinear i.e., O, S and M are at the same line.
Step 2
To prove O, S and M are collinear,
we have to prove
$\mathrm{△}FOS\sim \mathrm{△}CMS$
Since $CX\mathrm{\perp }AB$ and FY is the perpendicular bisector of AB. So we can say
$CX\mid \mid FY$
So alternate interior angle of transversal, when transversal intersect two parallel lines are congruent.
So
$\mathrm{\angle }SFO=\mathrm{\angle }SCM$
also we know that centroid S, splits the median into 2:1 ratio.
that is
CS=2FS
here M is the orthocenter of $\mathrm{△}ABC$ and O is the orthocenter of $\mathrm{△}DEF$.
and $\mathrm{△}ABC\sim \mathrm{△}DEF$ with 2:1 ratio.
So we have CM=2FO
Hence we can say
$\mathrm{△}FOS\sim \mathrm{△}CMS$ by SAS similarity.
So we can say
$\mathrm{\angle }FSO=\mathrm{\angle }CSM$
Hence we can say O, S and M are collinear and S divides the line segment MO in the ratio 1:2.