9. which statement must be true if $\overline{eg}$ is an altitude for the $\triangle def$?
image of triangle def with e at the top, d and f at the base, g on df, and eg drawn
$\overline{dg}$ is congruent to $\overline{fg}$.
$\triangle def$ is congruent to triangle $\triangle deg$.
$\triangle dge$ is congruent to $\triangle fge$.
$\overline{eg}$ is perpendicular to segment $\overline{df}$.

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Accepted Answer

# Explanation:
## Step1: Recall the definition of an altitude in a triangle.
An altitude of a triangle is a perpendicular segment from a vertex to the line containing the opposite side. So, if $\overline{EG}$ is an altitude of $\triangle DEF$, then $\overline{EG}$ must be perpendicular to $\overline{DF}$ (or the line containing $DF$).

## Step2: Analyze each option.
- Option 1: $\overline{DG}\cong\overline{FG}$ would be true if $EG$ is a median (or the triangle is isoceles with $ED = EF$), but altitude alone doesn't imply this.
- Option 2: $\triangle DEF\cong\triangle DEG$ is not true because $\triangle DEG$ is a part of $\triangle DEF$ and their sizes are different.
- Option 3: $\triangle DGE\cong\triangle FGE$ would require $DG = FG$ and $ED=EF$ (by HL or SAS), but altitude alone doesn't give congruence of these triangles.
- Option 4: By the definition of altitude, $\overline{EG}\perp\overline{DF}$ must be true.

# Answer:
The option " $\overline{EG}$ is perpendicular to segment $\overline{DF}$." (the last option)

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Question

Explanation:

Step1: Recall the definition of an altitude in a triangle.

Step2: Analyze each option.

Answer: