Talking Of Triangles

Given a triangle with rational coordinates, then all 3 sides are defined by lines with rational coefficients. All midpoints, being averages of coordinates of vertices, are also rational. All lines perpendicular to the sides passing through any vertex and any midpoints are also defined by lines with rational coefficients. Any intersection of any pair of lines described so far always has rational coordinates. This means the following points have rational coordinates:

A) Centroids, being the intersection of medians
B) Circumcenter, being the intersection of lines perpendicular to sides and passing through midpoints
C) Orthocenter, being the intersection of altitudes

This leaves the incenter, which is the intersection of angle bisectors. Consider the case where the base of the triangle has a slope of $0$ . Then the tangents of the two other sides are rational. But this means that the tangents of the angle bisectors are generally irrational, which means that the incenter generally doesn't have a rational coordindate.

$\tan\left( \dfrac { 1 }{ 2 } x \right) =\dfrac { \sqrt { 1+{ \left( \tan\left( x \right) \right) }^{ 2 } } -1 }{ \tan\left( x \right) }$

An easy counterexample is where the slopes of the sides of an isosceles are $\dfrac{4}{3}$ , which means that the slopes of the angle bisectors are $\dfrac{1}{2}$ , and thus the incenter will have a rational coordinate, so that sometimes all A, B, C, D can have rational coordinates, but not always.

As coordinates of the triangle are rational. Therefore centroid of the triangle must be rational.

And we know that centroid divides the line joining orthocentre and circumcentre in the ratio of 2:1... The two points must also be rational...

Use complex numbers. Shift all points to the unit circle. Then the center is 0 (rational). The Orthocenter is A+B+C which is rational and the Centroid is A/3 +B/3 + C/3 which is also rational. The incenter may not be rational because the roots of A,B,C may be irrational therefore making the incenter irrational.