Question

part 6 of 10
write a proof of the triangle midsegment theorem.
given: $overline{dg} \cong \overline{ge}$, $overline{fh} \cong \overline{he}$
prove: $overline{gh} \parallel \overline{df}$, $gh = \frac{1}{2}df$

b. $\angle edf \cong \angle efd$
c. $\angle egh \cong \angle edf$
d. $\angle egh \cong \angle efd$
by the converse of the corresponding angles theorem,

Explanation:

Step1: Identify similar triangles

We know $\overline{DG} \cong \overline{GE}$ and $\overline{FH} \cong \overline{HE}$, so $\frac{EG}{ED} = \frac{1}{2}$ and $\frac{EH}{EF} = \frac{1}{2}$. $\angle E$ is common to $\triangle EGH$ and $\triangle EDF$. By SAS similarity, $\triangle EGH \sim \triangle EDF$. Corresponding angles of similar triangles are congruent, so $\angle EGH \cong \angle EDF$.

Step2: Apply converse theorem

$\angle EGH$ and $\angle EDF$ are corresponding angles. By the Converse of the Corresponding Angles Theorem, if corresponding angles are congruent, the lines are parallel.

Answer:

$\overline{GH} \parallel \overline{DF}$

For the multiple-choice part, the correct option is:
C. $\angle EGH \cong \angle EDF$