Question

question 8 (multiple choice worth 1 points) (01.03r mc) if a translation maps ∠i onto ∠k, which of the following statements is true? diagram of a triangle with vertices g, h, i, j, k (h and i on the sides, j and k at the base) options: $overline{hi} cong overline{jk}$, $\triangle ghi sim \triangle gjk$, $overline{gh} cong overline{gj}$, $angle g cong angle k$

Explanation:

Step1: Recall Translation Properties

Translation is a rigid transformation, so it preserves shape, size, and angle measures, and maps segments to congruent segments and angles to congruent angles. Also, it creates similar figures (since congruent figures are similar with scale factor 1).

Step2: Analyze Each Option

- Option 1: $\overline{HI} \cong \overline{JK}$? Translation maps $\angle I$ to $\angle K$, so the direction of translation would relate $I$ to $K$ and $H$ to $J$. But $\overline{HI}$ and $\overline{JK}$: from the diagram, $\overline{HI}$ is a midline? Wait, no, translation maps $\angle I$ to $\angle K$, so vector from $I$ to $K$ is the translation vector. Then $H$ translates to $J$, $I$ to $K$, $G$ to some point? Wait, maybe better to look at triangles.
- Option 2: $\triangle GHI \sim \triangle GJK$? Since translation maps $\angle I$ to $\angle K$, and $HI \parallel JK$ (because translation preserves parallelism, as it's a rigid transformation and the direction of $HI$ and $JK$ would be same). So by AA similarity (since $\angle G$ is common, and $\angle GIH \cong \angle GKJ$ from translation), $\triangle GHI \sim \triangle GJK$. Wait, but actually, translation would make $\triangle GHI$ congruent to a triangle, but maybe the diagram shows $HI$ parallel to $JK$, so $\triangle GHI$ and $\triangle GJK$ are similar (since $HI \parallel JK$ implies corresponding angles equal, so AA similarity). Wait, but let's check other options.
- Option 3: $\overline{GH} \cong \overline{GJ}$? No, $GH$ is a segment from $G$ to $H$, $GJ$ is from $G$ to $J$. Unless $H$ and $J$ are same, which they aren't. So this is false.
- Option 4: $\angle G \cong \angle K$? $\angle G$ is at vertex $G$, $\angle K$ at $K$. Translation maps $\angle I$ to $\angle K$, so $\angle I \cong \angle K$, not $\angle G$. So this is false.

Wait, maybe I misread the options. Let's re-express:

First option: $\overline{HI} \cong \overline{JK}$? If translation maps $I$ to $K$ and $H$ to $J$, then $\overline{HI}$ translates to $\overline{JK}$, so $\overline{HI} \cong \overline{JK}$? Wait, but in the diagram, $HI$ is a midline? Wait, no, the triangle $GHI$ and $GJK$: if $HI$ is parallel to $JK$ (because translation preserves direction, so $HI$ and $JK$ are parallel), then by basic proportionality theorem (Thales' theorem), $HI$ is a midline, so $HI = \frac{1}{2}JK$, so $\overline{HI}
ot\cong \overline{JK}$. So first option is false.

Second option: $\triangle GHI \sim \triangle GJK$? Since $HI \parallel JK$ (from translation, as $I$ translates to $K$ and $H$ to $J$, so vector $IJ$ (wait, $H$ to $J$ and $I$ to $K$ are same vector, so $HJ \parallel IK$? Wait, no, translation vector is from $I$ to $K$, so $H$ translates to $J$ (since $H$ and $I$ are on the top triangle, $J$ and $K$ on the bottom). So $HI$ translates to $JK$, so $HI \parallel JK$ and $HI = JK$? Wait, no, translation preserves length, so if $I$ translates to $K$, and $H$ translates to $J$, then $\overline{HI} \cong \overline{JK}$, and $\angle I \cong \angle K$, $\angle H \cong \angle J$. Then $\triangle GHI$ and $\triangle GJK$: $\angle G$ is common, $\angle H \cong \angle J$, $\angle I \cong \angle K$, so they are similar (in fact, congruent? But maybe the diagram shows $GHI$ as a smaller triangle, so maybe similar with scale factor. Wait, maybe the correct option is $\triangle GHI \sim \triangle GJK$.

Wait, let's check the options again. The options are:

1. $\overline{HI} \cong \overline{JK}$

2. $\triangle GHI \sim \triangle GJK$

3. $\overline{GH} \cong \overline{GJ}$

4. $\angle G \cong \angle K$

So let's analyze each:

- Option 1: If translation maps $I$ to $K$ and $H$ to $J$, then $\overline{HI}$ translates to $\overline{JK}$, so $\overline{HI} \cong \overline{JK}$? But in the diagram, $HI$ is a horizontal segment (assuming) and $JK$ is also horizontal, same length? Wait, maybe. But let's check other options.

- Option 2: $\triangle GHI \sim \triangle GJK$: Since $HI \parallel JK$ (from translation, as $H$ translates to $J$ and $I$ to $K$, so the lines $HI$ and $JK$ are parallel), then by AA similarity ( $\angle G$ is common, and $\angle GHI \cong \angle GJK$ because $HI \parallel JK$ and $GJ$ is transversal, so corresponding angles equal; similarly $\angle GIH \cong \angle GKJ$), so $\triangle GHI \sim \triangle GJK$. This is true.

- Option 3: $\overline{GH} \cong \overline{GJ}$: $GH$ is from $G$ to $H$, $GJ$ is from $G$ to $J$. $H$ and $J$ are different points, so unless $GH = GJ$, which isn't implied by translation (translation maps $H$ to $J$, so $\overline{GH} \cong \overline{GJ}$? Wait, translation preserves length, so if $H$ translates to $J$, then $\overline{GH}$ translates to $\overline{GJ}$, so $\overline{GH} \cong \overline{GJ}$. Wait, maybe I made a mistake here.

Wait, translation vector is $\vec{v}$ such that $I + \vec{v} = K$ and $H + \vec{v} = J$. Then $G + \vec{v}$ would be some point, but $G$ is a vertex. Wait, maybe the triangle $GHI$ translates to triangle $G'JK$, but the problem says "a translation maps $\angle I$ onto $\angle K$", so the image of $\angle I$ is $\angle K$, so the vertex $I$ maps to $K$, and the sides of $\angle I$ ( $HI$ and $GI$ ) map to sides of $\angle K$ ( $JK$ and $GK$ ). So $HI$ maps to $JK$, so $\overline{HI} \cong \overline{JK}$, and $GI$ maps to $GK$, so $\overline{GI} \cong \overline{GK}$. Then $\triangle GHI$ and $\triangle GJK$: $HI \cong JK$, $GI \cong GK$, and $\angle I \cong \angle K$, so by SAS congruence, $\triangle GHI \cong \triangle GJK$, which implies they are similar (since congruent triangles are similar). But also, $\overline{GH}$ maps to $\overline{GJ}$ (since $H$ maps to $J$ and $G$ maps to $G$? Wait, no, translation would map $G$ to some point, but if $\angle I$ maps to $\angle K$, the center of translation? Wait, maybe the diagram is a triangle with a midline $HI$, so $H$ is midpoint of $GJ$ and $I$ is midpoint of $GK$. Then translation from $H$ to $J$ and $I$ to $K$ would have vector $HJ = IK$, so $HI \parallel JK$ and $HI = \frac{1}{2}JK$, so $\overline{HI}
ot\cong \overline{JK}$, so option 1 is false. Then $\triangle GHI$ and $\triangle GJK$: since $HI \parallel JK$, $\angle GHI = \angle GJK$ and $\angle GIH = \angle GKJ$, so by AA similarity, they are similar. So option 2 is true.

Option 3: $\overline{GH} \cong \overline{GJ}$: If $H$ is midpoint of $GJ$, then $GH = \frac{1}{2}GJ$, so $\overline{GH}
ot\cong \overline{GJ}$, so false.

Option 4: $\angle G \cong \angle K$: $\angle G$ is at $G$, $\angle K$ at $K$, no reason for them to be congruent, so false.

So the correct option is $\triangle GHI \sim \triangle GJK$.

Answer:

$\triangle GHI \sim \triangle GJK$ (the second option, assuming the options are labeled as, e.g., B. $\triangle GHI \sim \triangle GJK$)