(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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asked 2021-01-13
(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.

Answers (1)

2021-01-14
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 \(\displaystyle\triangle{A}{B}{C}\) and its medial triangle is \(\displaystyle\triangle{D}{E}{F}\)
image
Here O is orthocenter, M is circumcenter and S is barycenter.
Here we can say \(\displaystyle\triangle{A}{B}{C}\) is similar to \(\displaystyle\triangle{D}{E}{F}\) because D, E and F are the midpoint of lines BC, AC and AB respectively.
And also
\(\displaystyle{B}{C}{\mid}{\mid}{E}{F}\)
\(\displaystyle{A}{C}{\mid}{\mid}{F}{D}\)
\(\displaystyle{A}{B}{\mid}{\mid}{E}{D}\)
Hence
\(\displaystyle\triangle{A}{B}{C}\sim\triangle{D}{E}{F}\) with 2:1 ratio.
In the above figure O is the circumcenter of \(\displaystyle\triangle{A}{B}{C}\) which is also the orthocenter of \(\displaystyle\triangle{D}{E}{F}\).
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
\(\displaystyle\triangle{F}{O}{S}\sim\triangle{C}{M}{S}\)
Since \(\displaystyle{C}{X}\bot{A}{B}\) and FY is the perpendicular bisector of AB. So we can say
\(\displaystyle{C}{X}{\mid}{\mid}{F}{Y}\)
So alternate interior angle of transversal, when transversal intersect two parallel lines are congruent.
So
\(\displaystyle\angle{S}{F}{O}=\angle{S}{C}{M}\)
also we know that centroid S, splits the median into 2:1 ratio.
that is
CS=2FS
here M is the orthocenter of \(\displaystyle\triangle{A}{B}{C}\) and O is the orthocenter of \(\displaystyle\triangle{D}{E}{F}\).
and \(\displaystyle\triangle{A}{B}{C}\sim\triangle{D}{E}{F}\) with 2:1 ratio.
So we have CM=2FO
Hence we can say
\(\displaystyle\triangle{F}{O}{S}\sim\triangle{C}{M}{S}\) by SAS similarity.
So we can say
\(\displaystyle\angle{F}{S}{O}=\angle{C}{S}{M}\)
Hence we can say O, S and M are collinear and S divides the line segment MO in the ratio 1:2.
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