Question

part 4 of 10
write a proof of the triangle midsegment theorem.
given: $overline{dg} \cong \overline{ge}$, $overline{fh} \cong \overline{he}$
prove: $overline{gh} \parallel \overline{df}$, $gh = \frac{1}{2}df$

what is a known fact about similar triangles? select all that apply.
\\(\square\\) a. corresponding sides of similar triangles are proportional.
\\(\square\\) b. corresponding angles of similar triangles are not equal, but proportional.
\\(\square\\) c. corresponding angles of similar triangles are congruent.
\\(\square\\) d. corresponding sides of similar triangles are congruent.

Explanation:

• For option A: By definition, similar triangles have corresponding sides that are in proportional ratios.
• For option B: This is incorrect because corresponding angles of similar triangles are equal, not proportional.
• For option C: A key property of similar triangles is that all corresponding angles are congruent (equal).
• For option D: This describes congruent triangles, not similar triangles; similar triangles have proportional sides, not congruent ones.

Step1: Identify midpoints

Given $\overline{DG} \cong \overline{GE}$ and $\overline{FH} \cong \overline{HE}$, so $G$ is the midpoint of $\overline{DE}$, $H$ is the midpoint of $\overline{EF}$. Thus, $\frac{EG}{ED} = \frac{1}{2}$ and $\frac{EH}{EF} = \frac{1}{2}$.

Step2: Identify shared angle

$\angle GEH = \angle DEF$ (common angle to $\triangle GEH$ and $\triangle DEF$).

Step3: Prove triangle similarity

By SAS Similarity Criterion, since $\frac{EG}{ED} = \frac{EH}{EF}$ and included angles are congruent, $\triangle GEH \sim \triangle DEF$.

Step4: Deduce parallelism and side length

For similar triangles, corresponding angles are congruent: $\angle EGH = \angle EDF$, so $\overline{GH} \parallel \overline{DF}$ (corresponding angles converse). Also, $\frac{GH}{DF} = \frac{EG}{ED} = \frac{1}{2}$, so $GH = \frac{1}{2}DF$.

Answer:

A. Corresponding sides of similar triangles are proportional.
C. Corresponding angles of similar triangles are congruent.

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Proof of the Triangle Midsegment Theorem (included for completeness as part of the prompt):