From a point P in the plane of triangle ABC, perpendiculars are dropped to all three sides, each side being produced if the foot falls outside it. The three feet turn out to be collinear. Where must P lie?

The three feet are collinear exactly when P lies on the circumcircle, and the line they form is the Simson line of P. One easy check on the claim: take P to be the vertex A itself, which is certainly on the circumcircle — then two of the feet collapse onto A and the third is the foot of the altitude from A, so the three points are trivially in a line. The incircle has no such property; the orthocentre gives the three altitude feet, which form the orthic TRIANGLE and are definitely not collinear; and the line through the centroid and orthocentre is the Euler line, a different object altogether.