*my imagination*

* is a piece of board*

* my sole instrument*

* is a wooden stick*

*I strike the board*

* it answers me*

* yes—yes*

* no—no*

*“A Knocker” by Zbigniew Herbert*

Euler’s line theorem states

In every triangle

the intersection of the medians

the intersection of the heights

and the center of the circumscribed circle

are on a straight line

Let’s prove it

with the aid of Omega CAS Explorer

We know

is a circle centered at with radius

provide (2) is positive

**Find d, e, f from triangle’s vertices :**

*ceq:x^2+y^2+d*x+e*y+f=0;
eq1:ev(ceq, x=x1, y=0);
eq2:ev(ceq, x=-x1, y=0);
eq3:ev(ceq, x=x2, y=y2);
sol:linsolve([eq1, eq2, eq3], [d,e,f]);*

**Evaluate (2):**

*ev(d^2 + e^2-4*f, sol);*

always positive for

**Find the center of the circumscribed circle :**

*xc:ev(-d/2, sol)$*

yc:ev(-e/2, sol)$

**Find the intersection of the medians :**

*eq1:y*((x1+x2)/2+x1)=y2/2*(x+x1)$
eq2:y*((x2-x1)/2-x1)=y2/2*(x-x1)$
sol:solve([eq1, eq2], [x,y])$
xm:ev(x,sol);*

*ym:ev(y, sol);*

*is(ev(x2*y=y2*x, x=xm, y=ym));
*

**Find the intersection of the heights :**

*eq1:y2*y=-(x2+x1)*(x-x1)$
eq2:y2*y=-(x2-x1)*(x+x1)$
sol:solve([eq1, eq2], [x,y])$
xh:ev(x, sol);*

*yh:ev(y, sol);*

**Compute the area of triangle with vertices :**

*m:matrix([xc, yc,1], [xm, ym, 1], [xh, yh,1]);
determinant(m)$
ratsimp(%);*

Indeed

and are on a straight line.