These are so hard -- who keeps posting them? Let's go with a figure this time. A(5,3),B(7,8),C(2,1)
First, the easy endpoints are the midpoints of the sides:
D=(B+C)/2=(9/2,9/2)
E=(A+C)/2=(7/2,2)
F=(A+B)/2=(6,11/2)
Let's write the equations for the sides:
BC: (y-1)(7-2)=(x-2)(8-1) quad quad -7 x + 5 y = -9
AC: (y-1)(5-2)=(x-2)(3-1) quad quad -2 x + 3 y = - 1
AB: (y-3)(7-5)=(x-5)(8-3) quad quad -5 x + 2 y = - 19
For perpendicular bisectors we swap the coefficients on x and y and negate one. We get the constant by plugging in the midpoint.
DG bot BC: 5 x + 7y = 5(9/2) + 7(9/2)=54
EI bot AC: 3x + 2y = 3(7/2) + 2(2)=29/2
FH bot AB: 2 x + 5y = 2(6) + 5(11/2) = 79/2
G is the meet of DG and AB:
5 x + 7y=54
-5 x + 2 y = - 19
9y = 35
y = 35/9
x = 1/5 (54 - 7(35/9)) = 241/45
G = (241/45, 35/9)
DG=sqrt{(9/2 - 241/45)^2 + (9/2 - 35/9)^2} = {11 sqrt(74)}/90
Phew. That's one, though it may not be right.
I is the meet of EI with BC
3x + 2y =29/2
-7 x + 5 y = -9
30 x + 20 y =145
-28 x + 20 y = -36
58 x = 181
x= 181/58
y=149/58
I=(181/58, 149/58)
EI=sqrt{(7/2 - 181/58)^2 + (2 - 149/58)^2} = { 11 sqrt(13)}/58
FH meets BC at H
2 x + 5y = 79/2
-7 x + 5 y = -9
9 x = 97/2
x = 97/18
y = 517/90
H=(97/18, 517/90)
FH=sqrt((6 - 97/18)^2 + (11/2- 517/90)^2 ) = 11 sqrt{29}/90
