Euler line is a line passing through orthocenter H (blue), nine-point center N (red), centroid G (orange) and circumcenter O (green), while HN = NO and HG = 2GO. [1]
Put AB onto x-axis and C onto y-axis and set coordinates as A(-a, 0), B(b,0), C(0,c), where a, b and c are positive numbers, then we get all vertices and centers:
So it's easy to prove that HNGO are collinear and HN = NO and HG = 2GO.
The incenter, however, doesn't lie on the Euler line, unless the triangle is isosceles. We have:
Theorem 1 A triangle is isosceles if its incenter lies on its Euler line.
The proof (incenter lies on Euler line → isosceles) is not as easy as above because we should prove AB = AC or AB = BC or AC = BC, and exclude other possibilities.
Let's pick the centroid G, the orthocenter H (they are simpler than N and O) and the incenter I. The task is to proof that the determinant of collinearity has a form like:
where D = 0 iff AB = AC, E = 0 iff AB = BC, and F = 0 iff AC = BC.
If we choose above coordinates, the incenter will contain many square roots, which makes the determinant too difficult to be factored to D, E and F. So we have to run in a reverse way.
Given an incircle and two vertices A(-a,0) and B(b,0) on x-axis, then the third vertex C can be determined by two edges:
where AC and BC should be tangent to the incircle. Take the incircle and AC as example, eliminate x to get the quadratic equation about y, then we can solve k by setting discriminant to zero. The root k = 0 is edge AB and the non-zero root is edge AC. We use instead of
because the latter cannot cover the case that AC is parallel to y-axis.
A more simple way is to draw a circle orthogonal to the incircle with center A, such that two intersections are tangent points on AB (the origin) and AC. And we can get BC in the same way.
Now we have three vertices and three centers:
After some factoring work, we get the determinant of collinearity:
Obviously F = 0 iff AC = BC, but D and E are not obvious. So we should calculate the isosceles conditions:
which follows D = 0 iff AB = AC, E = 0 iff AB = BC, and F = 0 iff AC = BC.
Here is the computational proof process.
Here is another computational proof. More proofs can be found here.
- We use the diagram from here.