### note: close this window to return to Amy's Write-Up
#4

## 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**.

**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.**