1. **State the problem:** We have parallelogram ABCD reflected across the x-axis and then rotated 180° clockwise about the origin to form parallelogram EFGH. 2. **Recall transformations:** - Reflection across the x-axis changes a point $(x,y)$ to $(x,-y)$. - Rotation 180° clockwise about the origin changes $(x,y)$ to $(-x,-y)$. 3. **Find coordinates of EFGH:** Given vertices: $A(-5,4), B(-1,4), C(-3,1), D(-7,1)$ Reflect across x-axis: $A'(-5,-4), B'(-1,-4), C'(-3,-1), D'(-7,-1)$ Rotate 180° clockwise: $E = (-(-5), -(-4)) = (5,4)$ $F = (-(-1), -(-4)) = (1,4)$ $G = (-(-3), -(-1)) = (3,1)$ $H = (-(-7), -(-1)) = (7,1)$ 4. **Check parallel sides:** - Side EF vector: $F - E = (1-5, 4-4) = (-4,0)$ - Side GH vector: $H - G = (7-3, 1-1) = (4,0)$ Vectors $(-4,0)$ and $(4,0)$ are parallel (same direction or opposite). - Side EH vector: $H - E = (7-5, 1-4) = (2,-3)$ - Side FG vector: $G - F = (3-1, 1-4) = (2,-3)$ Vectors $(2,-3)$ and $(2,-3)$ are parallel. 5. **Conclusion:** - Side EF is parallel to side GH: **True** - Side EH is parallel to side FG: **True**