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.
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.
equation of line containing AD:
easy! x = d.
Now, we are given that AD and CH intersect at point E, so let's find the coordinates of E. The x-coordinate is obviously d. Substitute into y = -dx/a + dc/a to find that the y-coordinate of E is (-d^2 + dc)/a.
Finally, check to see if this point E (d, (-d^2 + dc)/a) lies on BG by plugging in x = d.
y = (c - d)d/a = (cd - d^2)/a = (-d^2 + cd)/a = y-coordinate of E. Ta da!
So, E must lie on BG, and therefore we have shown that all three lines containing the altitudes intersect at point E.
*remember to close this window to return to Amy's Write-Up #4.