To solve the problem, we need to find the expression \(\sqrt{|AG|^2 + |BG|^2 - |CG|^2}\) where \(G\) is the centroid of triangle \(ABC\) and \(D(-2, 3)\), \(E(4, -3)\), and \(F(4, 5)\) are the midpoints of sides \(BC\), \(CA\), and \(AB\) respectively.
### Step-by-Step Solution
1. **Identify Coordinates of Midpoints**:
- The coordinates of midpoints are given as:
- \(D(-2, 3)\)
- \(E(4, -3)\)
- \(F(4, 5)\)
2. **Set Up Equations for Vertices**:
- Let the coordinates of vertices \(A\), \(B\), and \(C\) be \(A(x_1, y_1)\), \(B(x_2, y_2)\), \(C(x_3, y_3)\).
- From the midpoint \(D\):
\[
D = \left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}\right) = (-2, 3)
\]
This gives us two equations:
\[
\frac{x_2 + x_3}{2} = -2 \quad \Rightarrow \quad x_2 + x_3 = -4 \quad (1)
\]
\[
\frac{y_2 + y_3}{2} = 3 \quad \Rightarrow \quad y_2 + y_3 = 6 \quad (2)
\]
- From the midpoint \(E\):
\[
E = \left(\frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}\right) = (4, -3)
\]
This gives us:
\[
\frac{x_1 + x_3}{2} = 4 \quad \Rightarrow \quad x_1 + x_3 = 8 \quad (3)
\]
\[
\frac{y_1 + y_3}{2} = -3 \quad \Rightarrow \quad y_1 + y_3 = -6 \quad (4)
\]
- From the midpoint \(F\):
\[
F = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = (4, 5)
\]
This gives us:
\[
\frac{x_1 + x_2}{2} = 4 \quad \Rightarrow \quad x_1 + x_2 = 8 \quad (5)
\]
\[
\frac{y_1 + y_2}{2} = 5 \quad \Rightarrow \quad y_1 + y_2 = 10 \quad (6)
\]
3. **Solve the System of Equations**:
- From equations (1) and (3):
\[
x_2 + x_3 = -4 \quad (1)
\]
\[
x_1 + x_3 = 8 \quad (3)
\]
Substituting \(x_3 = -4 - x_2\) into (3):
\[
x_1 - 4 - x_2 = 8 \quad \Rightarrow \quad x_1 - x_2 = 12 \quad (7)
\]
- From equations (5) and (7):
\[
x_1 + x_2 = 8 \quad (5)
\]
Adding (5) and (7):
\[
2x_1 = 20 \quad \Rightarrow \quad x_1 = 10
\]
Substituting back to find \(x_2\):
\[
10 + x_2 = 8 \quad \Rightarrow \quad x_2 = -2
\]
Using (1) to find \(x_3\):
\[
-2 + x_3 = -4 \quad \Rightarrow \quad x_3 = -2
\]
- Now for \(y\) coordinates using (2), (4), and (6):
\[
y_2 + y_3 = 6 \quad (2)
\]
\[
y_1 + y_3 = -6 \quad (4)
\]
Substitute \(y_3 = 6 - y_2\) into (4):
\[
y_1 + 6 - y_2 = -6 \quad \Rightarrow \quad y_1 - y_2 = -12 \quad (8)
\]
From (6):
\[
y_1 + y_2 = 10 \quad (6)
\]
Adding (6) and (8):
\[
2y_1 = -2 \quad \Rightarrow \quad y_1 = -1
\]
Substituting back to find \(y_2\):
\[
-1 + y_2 = 10 \quad \Rightarrow \quad y_2 = 11
\]
Using (2) to find \(y_3\):
\[
11 + y_3 = 6 \quad \Rightarrow \quad y_3 = -5
\]
4. **Coordinates of Vertices**:
- The coordinates of the vertices are:
- \(A(10, -1)\)
- \(B(-2, 11)\)
- \(C(-2, -5)\)
5. **Find the Centroid \(G\)**:
- The coordinates of the centroid \(G\) are given by:
\[
G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) = \left(\frac{10 - 2 - 2}{3}, \frac{-1 + 11 - 5}{3}\right) = \left(\frac{6}{3}, \frac{5}{3}\right) = (2, \frac{5}{3})
\]
6. **Calculate Distances**:
- Calculate \(AG\), \(BG\), and \(CG\):
\[
|AG| = \sqrt{(10 - 2)^2 + \left(-1 - \frac{5}{3}\right)^2} = \sqrt{8^2 + \left(-\frac{8}{3}\right)^2} = \sqrt{64 + \frac{64}{9}} = \sqrt{\frac{640}{9}} = \frac{8\sqrt{10}}{3}
\]
\[
|BG| = \sqrt{(-2 - 2)^2 + \left(11 - \frac{5}{3}\right)^2} = \sqrt{(-4)^2 + \left(\frac{28}{3}\right)^2} = \sqrt{16 + \frac{784}{9}} = \sqrt{\frac{144 + 784}{9}} = \sqrt{\frac{928}{9}} = \frac{4\sqrt{58}}{3}
\]
\[
|CG| = \sqrt{(-2 - 2)^2 + \left(-5 - \frac{5}{3}\right)^2} = \sqrt{(-4)^2 + \left(-\frac{20}{3}\right)^2} = \sqrt{16 + \frac{400}{9}} = \sqrt{\frac{144 + 400}{9}} = \sqrt{\frac{544}{9}} = \frac{4\sqrt{34}}{3}
\]
7. **Final Calculation**:
- Now substitute into the expression:
\[
\sqrt{|AG|^2 + |BG|^2 - |CG|^2} = \sqrt{\left(\frac{8\sqrt{10}}{3}\right)^2 + \left(\frac{4\sqrt{58}}{3}\right)^2 - \left(\frac{4\sqrt{34}}{3}\right)^2}
\]
\[
= \sqrt{\frac{64 \cdot 10 + 16 \cdot 58 - 16 \cdot 34}{9}} = \sqrt{\frac{640 + 928 - 544}{9}} = \sqrt{\frac{1024}{9}} = \frac{32}{3}
\]
Thus, the final answer is:
\[
\frac{32}{3}
\]
