According to the midsegment theorem, the midsegments are parallel to
and half the length of the opposite side.
The completed statement are as follows;
Reasons:
From the given graph of ΔABC, we have;
Coordinates of the points A, B, and C are; A(-5, 2), B(1, -2), and C(-3, -6)
The coordinates of the point D and E on [tex]\mathbf{\overline{DE}}[/tex] are; D(-4, -2), and E(-2, 0)
The coordinates of the point F is; F(-1, -4)
[tex]\displaystyle Slope \ of \ line \ \overline{AC} = \mathbf{\frac{(-6) - 2}{-3 - (-5)} = \frac{-8}{2}} = -4[/tex]
[tex]\displaystyle Slope \ of \ line \ \overline{EF} = \frac{0 - (-4)}{-2 - (-1)} = \frac{4}{-1} = -4[/tex]
Length of EF = √((-1 - (-2))² + (-4 - 0)²) = √(17)
Length of AC = √((-3 - (-5))² + (-6 - 2)²) = √(4 × 17) = 2·√(17)
Therefore, we have;
Because the slope of [tex]\mathbf{\overline {EF}}[/tex] and [tex]\mathbf{\overline {AC}}[/tex] are both , -4, [tex]\overline {EF}[/tex] ║ [tex]\overline {AC}[/tex]. EF = [tex]\underline{\sqrt{17}}[/tex], and AC
= [tex]\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17}}[/tex],. Because [tex]\underline{\sqrt{17}}[/tex] = [tex]\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17}}[/tex], [tex]EF =\mathbf{ \frac{1}{2} AC}[/tex]
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