## Proof that the lines containing the altitudes of any triangle intersect in one point

Given: triangle ABC has coordinates as shown (conveniently placed on coordinate axes; note that placing B at the origin and letting segment BC coincide with the x-axis does not lose generality.)
AD perpendicular to BC; CH perpendicular to AB; BG perpendicular to AC.
AD and CH intersect at E

Prove: BG passes through point E.

Proof:

First let's find the equations for the lines containing the altitudes of triangle ABC in terms of the coordinates already given.

equation of line containing CH:

Note that the equation for line CH has a slope that is the opposite reciprocal of the slope of line AB.

Slope of AB = a/d, so slope of CH = -d/a.

Since (c, 0) lies on line CH, the equation for that line is y - 0 = -d/a(x - c), or y = -dx/a + dc/a.

equation of line containing BG:

Slope of AC = a/(d - c), so slope of BG = (c - d)/a.

Since (0, 0) lies on line BG, the equation for that line is y = (c - d)x/a.