Question

write a proof of the triangle midsegment theorem. given: \\(\overline{dg} \cong \overline{ge}\\), \\(\overline{fh} \cong \overline{he}\\) prove: \\(\overline{gh}\parallel\overline{df}\\), \\(gh = \frac{1}{2}df\\) what is \\(\frac{eh}{ef}\\)? \\(\frac{eh}{ef} = \square\\) (type an integer or a simplified fraction.)

Explanation:

Step1: Identify segment congruence

Given $\overline{FH} \cong \overline{HE}$, so $EH = FH$.

Step2: Express EF in terms of EH

$\overline{EF} = \overline{EH} + \overline{FH} = \overline{EH} + \overline{EH} = 2\overline{EH}$

Step3: Calculate the ratio

Substitute into $\frac{EH}{EF}$:
$\frac{EH}{EF} = \frac{EH}{2EH} = \frac{1}{2}$

Answer:

$\frac{1}{2}$