Answer
The perpendicular bisectors are concurrent, for they all intersect at a common point for a triangle.
Work Step by Step
To prove that the perpendicular bisectors 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 (2b,2c), and C the point (2a,0). We find the equation of a line perpendicular to BC:
$m_{BC}= \frac{2c-0}{2b-2a} = \frac{c}{b-a}$
The equation of the perpendicular bisector is:
$y - c =\frac{b-c}{-c} (x -a-b)$
The slope of the bisector is the inverse reciprocal of:
$m_{AC} = \frac{c-0}{b-0}=\frac{c}{b}$
Thus:
$y - c=\frac{-b}{c}(x-b)$
We now see where these lines intersect:
$\frac{-b}{c}(x-b) = \frac{b-c}{-c} (x -a-b) \\ x-b = \frac{b-c}{b}(x-a-b) \\ x-b =\frac{b-c}{b}x - \frac{b-c}{b}a -b+c\\ x -\frac{b-c}{b}x=- \frac{b-c}{b}a+c \\ \frac{c}{b} x = \frac{-ba+ca+cb}{b} \\ x = a$
a is the midpoint of the first line, which is where the vertical perpendicular bisector goes through, so we see that they all intersect at a common point.
