To find the coordinates of points $A$, $B$, and $C$ given that $D(7,3)$, $E(19,9)$, and $F(5,5)$ are the midpoints of sides $AB$, $BC$, and $AC$ of triangle $ABC$, we can use the midpoint formula.

Step 1: Set up the midpoint equations

Let's denote the coordinates of $A$ as $(x_1, y_1)$, $B$ as $(x_2, y_2)$, and $C$ as $(x_3, y_3)$.

The midpoint of a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

Using this formula for each midpoint:

1. For $D(7,3)$, the midpoint of $AB$:

   $D = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = (7,3)$

   This gives us the equations: $\frac{x_1 + x_2}{2} = 7 \quad \text{and} \quad \frac{y_1 + y_2}{2} = 3$ Simplifying: $x_1 + x_2 = 14 \quad \text{and} \quad y_1 + y_2 = 6$

2. For $E(19,9)$, the midpoint of $BC$:

   $E = \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right) = (19,9)$

   This gives us the equations: $\frac{x_2 + x_3}{2} = 19 \quad \text{and} \quad \frac{y_2 + y_3}{2} = 9$ Simplifying: $x_2 + x_3 = 38 \quad \text{and} \quad y_2 + y_3 = 18$

3. For $F(5,5)$, the midpoint of $AC$:

   $F = \left( \frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2} \right) = (5,5)$

   This gives us the equations: $\frac{x_1 + x_3}{2} = 5 \quad \text{and} \quad \frac{y_1 + y_3}{2} = 5$ Simplifying: $x_1 + x_3 = 10 \quad \text{and} \quad y_1 + y_3 = 10$

Step 2: Solve the system of equations

Now we have three pairs of linear equations:

(1) \quad & x_1 + x_2 = 14 \quad & (2) \quad & y_1 + y_2 = 6 \\ (3) \quad & x_2 + x_3 = 38 \quad & (4) \quad & y_2 + y_3 = 18 \\ (5) \quad & x_1 + x_3 = 10 \quad & (6) \quad & y_1 + y_3 = 10 \\ \end{align*}$$ #### Solving for $$x_1$$, $$x_2$$, and $$x_3$$: 1. From equations (1) and (5): $$\begin{align*} x_1 + x_2 &= 14 \quad & \text{(Equation 1)} \\ x_1 + x_3 &= 10 \quad & \text{(Equation 5)} \\ \end{align*}$$ Subtract (5) from (1): \[ x_2 - x_3 = 4 \quad \text{(Equation 7)}