(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 · Accepted Answer

The line on which orthocenter, circumcenter and barycenter(centroid) lie is called Euler Line of the triangle.Let a triangle&amp;nbsp;△ABC&amp;nbsp;and its medial triangle is&amp;nbsp;△DEFO is orthocenter, M is circumcenter and S is barycenter.△ABC&amp;nbsp;is similar to&amp;nbsp;△DEF&amp;nbsp;because D, E and F are the midpoint of lines BC, AC and AB respectively, soBC∣∣EFAC∣∣FDAB∣∣EDSo,&amp;nbsp;△ABC∼△DEF&amp;nbsp;with 2:1 ratio.In the above figure O is the circumcenter of&amp;nbsp;△ABC&amp;nbsp;which is also the orthocenter of&amp;nbsp;△DEF.Now we have to prove O, S and M are collinear i.e., O, S and M are at the same line.To prove O, S and M are collinear,we have to prove△FOS∼△CMSSince&amp;nbsp;CX⊥AB&amp;nbsp;and FY is the perpendicular bisector of AB. So we can sayCX∣∣FYSo alternate interior angle of transversal, when transversal intersect two parallel lines are congruent.So,&amp;nbsp;∠SFO=∠SCMalso we know that centroid S, splits the median into 2:1 ratio.that is CS=2FShere M is the orthocenter of&amp;nbsp;△ABC&amp;nbsp;and O is the orthocenter of&amp;nbsp;△DEF.and&amp;nbsp;△ABC∼△DEF&amp;nbsp;with 2:1 ratio.So we have CM=2FOSo&amp;nbsp;△FOS∼△CMS&amp;nbsp;by SAS similarity.So&amp;nbsp;∠FSO=∠CSMSo, O, S and M are collinear and S divides the line segment MO 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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