#### Answer

For a triangle, the altitudes are always concurrent.

#### Work Step by Step

To prove that the altitudes are concurrent, we must prove that they intersect at the same point. We first define coordinates, calling A the point (0,0), B the point (b,c), and C the point (a,0). We find the equation of a line perpendicular to BC:
$m_{BC}= \frac{c-0}{b-c} = \frac{c}{b-c}$
The altitude is perpendicular to this line an goes through (0,0), so we find that its equation is:
$y =\frac{a-b}{c} x$
The slope of the other altitude is the inverse reciprocal of:
$m_{AC} = \frac{c-0}{b-0}=\frac{c}{b}$
Thus:
$y=\frac{-b}{c}(x-a)$
We now see where these lines intersect:
$\frac{a-b}{c} x =\frac{-b}{c}(x-a) \\\frac{a-b}{-b} x = x-a \\ \frac{-a+b}{b} x - x = -a \\ \frac{-a+b-b}{b} x = -a \\ x = b$
Since the other altitude is the vertical line $x=b$, it follows that the altitudes of a triangle are concurrent.