1. **State the problem:** We are given triangle $\triangle ABC$ with midsegments $DE$, $EF$, and $DF$. We need to find the coordinates of points $D$, $E$, and $F$ (the midpoints of the sides), and then show that each midsegment is parallel to the opposite side and half its length. 2. **Recall the Midsegment Theorem:** The midsegment connecting the midpoints of two sides of a triangle is parallel to the third side and its length is half the length of that side. 3. **Find coordinates of $D$, $E$, and $F$: ** - $D$ is midpoint of $AB$: If $A=(x_A,y_A)$ and $B=(x_B,y_B)$, then $$D=\left(\frac{x_A+x_B}{2},\frac{y_A+y_B}{2}\right)$$ - $E$ is midpoint of $BC$: If $B=(x_B,y_B)$ and $C=(x_C,y_C)$, then $$E=\left(\frac{x_B+x_C}{2},\frac{y_B+y_C}{2}\right)$$ - $F$ is midpoint of $AC$: If $A=(x_A,y_A)$ and $C=(x_C,y_C)$, then $$F=\left(\frac{x_A+x_C}{2},\frac{y_A+y_C}{2}\right)$$ 4. **Show $DE \parallel CB$ and $DE=\frac{1}{2}CB$: ** - Vector $DE = E - D$ - Vector $CB = B - C$ - Show $DE$ is scalar multiple of $CB$ with factor $\frac{1}{2}$ 5. **Show $EF \parallel AC$ and $EF=\frac{1}{2}AC$: ** - Vector $EF = F - E$ - Vector $AC = C - A$ - Show $EF$ is scalar multiple of $AC$ with factor $\frac{1}{2}$ 6. **Show $DF \parallel AB$ and $DF=\frac{1}{2}AB$: ** - Vector $DF = F - D$ - Vector $AB = B - A$ - Show $DF$ is scalar multiple of $AB$ with factor $\frac{1}{2}$ **Note:** Since the exact coordinates of $A$, $B$, and $C$ are not provided in the text, we cannot compute numeric answers here. The method above applies once coordinates are known. **Final answer:** The midsegments $DE$, $EF$, and $DF$ connect midpoints of sides of $\triangle ABC$, are parallel to the opposite sides, and have lengths half of those sides, confirming the Triangle Midsegment Theorem.