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 \(\displaystyle\triangle{D}{E}{F}\)

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.

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 \(\displaystyle\triangle{D}{E}{F}\)

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.