Reflections of a Line Through the Orthocenter: What is this about?
A Mathematical Droodle

Explanation

The following result is due to L. Carnot (1801):

The problem quite easily solved by "angle chasing" - counting angles which, due to the layout of the problem, come in just several related sizes. However, configurations differ in details and several cases ought to be considered. That said, the idea of the proof can be introduced with a particular case:

Assume a line through the orthocenter H of ABC crosses AC at T_b and BC at T_a. Denote angle CT_bH as b and angle BT_aH as a. Then

(1)	a + b + ACB = 180o.

As we know, reflection K_a of the orthocenter H in BC lies on the circumcircle, as does its reflection K_b in AC. Next observe that the lines T_aK_a and T_bK_b are exactly the reflections of the given line in BC and AC, respectively. Assume the two meet in Q and let's show that Q lies on the circumcircle of the given triangle.

In QT_aT angle T_a is obviously 180^o - 2a, while angle T_b measures 180^o - 2b. Thus

TaQTb = 180o - (180o - 2a) - (180o - 2b)   = 2(a + b) - 180o   = 2(180o - ACB) - 180o   = 2(90o - ACB),

where we applied (1) on the last step. But this is exactly half the angular measure of arc K_aCK_b that subtends the angle K_aQK_b. The latter is therefore inscribed into the circumcircle, so that indeed this is where Q lies and T_aK_a and T_bK_b meet.

There are two more pairs of reflections to consider, but since each reflection participates in two pairs but intersects the circle only once (not counting the points K), all three pairs meet in the same point which is bound to be Q.