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Accepted Answer

# Explanation:
## Step1: Recall Hinge Theorem Converse
The Converse of the Hinge Theorem states that if two sides of one triangle are congruent to two sides of another triangle, and the third side of one triangle is longer than the third side of the other, then the included angle opposite the longer third side is larger.

## Step2: Identify Congruent Sides and Included Angles
We have $\overline{HI} \cong \overline{KL}$, $\overline{GH} \cong \overline{JK}$, and $\overline{JL} > \overline{GI}$. The included angles for the sides $\overline{GH}$ and $\overline{HI}$ (in $\triangle GHI$) is $\angle H$, and for $\overline{JK}$ and $\overline{KL}$ (in $\triangle JKL$) is $\angle K$? Wait, no—wait, the sides are $\overline{GH} \cong \overline{JK}$, $\overline{HI} \cong \overline{KL}$, so the included angles are $\angle G$ (between $\overline{GH}$ and $\overline{GI}$) and $\angle J$ (between $\overline{JK}$ and $\overline{JL}$)? Wait, no, let's re-express the triangles.

In $\triangle GHI$: sides $\overline{GH}$, $\overline{HI}$, and $\overline{GI}$. In $\triangle JKL$: sides $\overline{JK}$, $\overline{KL}$, and $\overline{JL}$. Given $\overline{GH} \cong \overline{JK}$, $\overline{HI} \cong \overline{KL}$, and $\overline{JL} > \overline{GI}$. So the two sides $\overline{GH} \cong \overline{JK}$ and $\overline{HI} \cong \overline{KL}$, so the included angles are $\angle G$ (between $\overline{GH}$ and $\overline{GI}$)? No, wait, $\overline{HI}$ is a side, so in $\triangle GHI$, the sides adjacent to $\angle G$ are $\overline{GH}$ and $\overline{GI}$? No, $\overline{HI}$ is the base. Wait, maybe I mixed up. Let's list the sides:

For $\triangle GHI$: sides $GH$, $HI$, $GI$.

For $\triangle JKL$: sides $JK$, $KL$, $JL$.

Given $GH \cong JK$, $HI \cong KL$, and $JL > GI$.

The included angles for the two congruent sides ($GH \cong JK$ and $HI \cong KL$) are $\angle H$ (in $\triangle GHI$, between $GH$ and $HI$) and $\angle K$ (in $\triangle JKL$, between $JK$ and $KL$)? No, that doesn't fit. Wait, no, the Hinge Theorem (and its converse) deals with two sides and the included angle. So if two sides of one triangle are congruent to two sides of another, then the larger third side implies a larger included angle.

So in $\triangle GHI$: sides $GH$, $HI$, $GI$. In $\triangle JKL$: sides $JK$, $KL$, $JL$. $GH \cong JK$, $HI \cong KL$, $JL > GI$. So the two sides $GH$ and $HI$ in $\triangle GHI$ are congruent to $JK$ and $KL$ in $\triangle JKL$. The included angle for $GH$ and $HI$ is $\angle H$, and for $JK$ and $KL$ is $\angle K$? No, that can't be. Wait, maybe the triangles are $\triangle GHI$ with sides $GH$, $GI$, $HI$ and $\triangle JKL$ with sides $JK$, $JL$, $KL$. So $GH \cong JK$, $HI \cong KL$, and $JL > GI$. Then the included angles are $\angle G$ (between $GH$ and $GI$) and $\angle J$ (between $JK$ and $JL$). So by the Converse of Hinge Theorem, since $JL > GI$, the included angle $\angle J$ (opposite $JL$?) Wait, no, the included angle is between the two congruent sides. So $GH \cong JK$ and $GI$ vs $JL$. Wait, maybe I need to re-express:

Let’s denote:

- In $\triangle GHI$: $a = GH$, $b = HI$, $c = GI$
- In $\triangle JKL$: $a' = JK$, $b' = KL$, $c' = JL$

Given $a = a'$, $b = b'$, and $c' > c$. By Converse Hinge Theorem, the included angle for $a$ and $b$ in $\triangle GHI$ is $\angle H$, and in $\triangle JKL$ is $\angle K$? No, that's not right. Wait, the included angle is between the two sides that are congruent. So $a = a'$ (GH = JK) and $b = b'$ (HI = KL), so the included angle is between $a$ and $b$, which is $\angle H$ (in $\triangle GHI$) and $\angle K$ (in $\triangle JKL$)? But then if $c' > c$, then the included angle $\angle K$ should be larger than $\angle H$? Wait, no, maybe I got the triangles wrong.

Wait, the sides are $GH \cong JK$, $HI \cong KL$, so the two sides are $GH$ and $HI$ in $\triangle GHI$, and $JK$ and $KL$ in $\triangle JKL$. So the included angles are $\angle G$ (between $GH$ and $GI$)? No, $GI$ is the third side. Wait, maybe the triangles are $\triangle GHI$ with vertices $G$, $H$, $I$, so sides $GH$, $HI$, $GI$. $\triangle JKL$ with vertices $J$, $K$, $L$, sides $JK$, $KL$, $JL$. So $GH \cong JK$, $HI \cong KL$, so the two sides are $GH$ and $HI$ (in $\triangle GHI$) and $JK$ and $KL$ (in $\triangle JKL$). The included angle for $GH$ and $HI$ is $\angle H$, and for $JK$ and $KL$ is $\angle K$. But then the third sides are $GI$ and $JL$. Wait, but $GI$ is between $G$ and $I$, and $JL$ is between $J$ and $L$. So maybe the included angles are $\angle G$ and $\angle J$? Wait, I think I made a mistake. Let's check the Hinge Theorem again.

The Hinge Theorem (SAS Inequality) states that if two sides of one triangle are congruent to two sides of another triangle, and the included angle of the first is larger than the included angle of the second, then the third side of the first is longer than the third side of the second. The Converse (SSS Inequality) states that if two sides of one triangle are congruent to two sides of another triangle, and the third side of the first is longer than the third side of the second, then the included angle of the first is larger than the included angle of the second.

So in our case:

- Two sides congruent: $GH \cong JK$ and $HI \cong KL$ (so $a = a'$, $b = b'$)
- Third side: $JL > GI$ (so $c' > c$)

Therefore, the included angle for $a$ and $b$ in $\triangle JKL$ (where the third side is longer) should be larger than the included angle in $\triangle GHI$.

Now, which angle is the included angle for $GH$ and $HI$ in $\triangle GHI$? That's $\angle H$ (between $GH$ and $HI$).

Which angle is the included angle for $JK$ and $KL$ in $\triangle JKL$? That's $\angle K$ (between $JK$ and $KL$)? Wait, no, that can't be, because $JL$ is the third side, not $KL$. Wait, maybe the sides are $GH \cong JK$, $GI \cong...$ No, the problem says $\overline{HI} \cong \overline{KL}$, $\overline{GH} \cong \overline{JK}$, $\overline{JL} > \overline{GI}$. So the two sides are $GH$ (side 1) and $HI$ (side 2) in $\triangle GHI$, and $JK$ (side 1) and $KL$ (side 2) in $\triangle JKL$. The third sides are $GI$ (in $\triangle GHI$) and $JL$ (in $\triangle JKL$). So the included angles are $\angle G$ (between $GH$ and $GI$) and $\angle J$ (between $JK$ and $JL$)? Wait, no, $GH$ and $HI$ are the two congruent sides, so the included angle is between $GH$ and $HI$, which is $\angle H$, and between $JK$ and $KL$, which is $\angle K$. But then the third sides are $GI$ and $JL$, which are opposite $\angle H$ and $\angle K$? No, in a triangle, the side opposite an angle is across from it. So in $\triangle GHI$, side $GI$ is opposite $\angle H$, and in $\triangle JKL$, side $JL$ is opposite $\angle K$. Wait, that makes sense! So if $JL > GI$, then the angle opposite $JL$ (which is $\angle K$) should be larger than the angle opposite $GI$ (which is $\angle H$)? No, wait, no: in $\triangle GHI$, side $GI$ is opposite $\angle H$; in $\triangle JKL$, side $JL$ is opposite $\angle K$. So if $JL > GI$, then by the Converse of Hinge Theorem (which is similar to SSS Inequality), the angle opposite the longer side is larger. Wait, but Hinge Theorem is about two sides and included angle. Wait, maybe I confused the theorem.

Wait, let's re-express the triangles with correct notation. Let’s define:

- $\triangle GHI$: vertices $G$, $H$, $I$. Sides: $GH$, $HI$, $GI$.
- $\triangle JKL$: vertices $J$, $K$, $L$. Sides: $JK$, $KL$, $JL$.

Given:

- $GH \cong JK$ (side 1)
- $HI \cong KL$ (side 2)
- $JL > GI$ (side 3)

By the Converse of the Hinge Theorem, since two sides of $\triangle JKL$ ( $JK$ and $KL$ ) are congruent to two sides of $\triangle GHI$ ( $GH$ and $HI$ ), and the third side $JL$ of $\triangle JKL$ is longer than the third side $GI$ of $\triangle GHI$, then the included angle between the two congruent sides in $\triangle JKL$ is larger than the included angle in $\triangle GHI$.

The included angle between $JK$ and $KL$ in $\triangle JKL$ is $\angle K$, and the included angle between $GH$ and $HI$ in $\triangle GHI$ is $\angle H$? No, that's not right, because $JL$ is not between $JK$ and $KL$. Wait, I think the mistake is in identifying the included angle. The two congruent sides are $GH$ and $JK$ (one pair), and $HI$ and $KL$ (another pair). So the included angles are between $GH$ and $HI$ (which is $\angle H$) and between $JK$ and $KL$ (which is $\angle K$). But then the third sides are $GI$ (opposite $\angle H$) and $JL$ (opposite $\angle K$). So if $JL > GI$, then $\angle K > \angle H$? But the options are $m\angle J$, $m\angle H$, $m\angle G$, $m\angle K$.

Wait, maybe the triangles are $\triangle GHI$ with $GH = JK$, $GI$ as a side, and $\triangle JKL$ with $JK = GH$, $JL$ as a side, and $HI = KL$. So the two sides are $GH$ and $GI$ in $\triangle GHI$, and $JK$ and $JL$ in $\triangle JKL$, with $HI = KL$. Wait, the problem states $\overline{HI} \cong \overline{KL}$, so $HI = KL$, $\overline{GH} \cong \overline{JK}$ (so $GH = JK$), and $\overline{JL} > \overline{GI}$ (so $JL > GI$). So in $\triangle GHI$: sides $GH$, $HI$, $GI$. In $\triangle JKL$: sides $JK$, $KL$, $JL$. So $GH = JK$, $HI = KL$, $JL > GI$. So the two sides $GH$ and $HI$ in $\triangle GHI$ are equal to $JK$ and $KL$ in $\triangle JKL$, and the third side $JL$ is longer than $GI$. Therefore, the included angle for $GH$ and $HI$ in $\triangle GHI$ is $\angle G$ (between $GH$ and $GI$)? No, $\angle G$ is between $GH$ and $GI$, but $HI$ is the base. Wait, I think I need to look at the vertices. Let's list the angles:

- In $\triangle GHI$: angles are $\angle G$ (at $G$), $\angle H$ (at $H$), $\angle I$ (at $I$)
- In $\triangle JKL$: angles are $\angle J$ (at $J$), $\angle K$ (at $K$), $\angle L$ (at $L$)

Given $GH \cong JK$, $HI \cong KL$, so the sides adjacent to $\angle G$ are $GH$ and $GI$, and adjacent to $\angle J$ are $JK$ and $JL$. Wait, maybe the included angles are $\angle G$ and $\angle J$, because $GH \cong JK$ and $GI$ vs $JL$. So if $JL > GI$, then by Converse Hinge Theorem, $\angle J > \angle G$, because $\angle J$ is the included angle between $JK$ and $JL$, and $\angle G$ is the included angle between $GH$ and $GI$, with $JK \cong GH$ and $JL > GI$.

Yes, that makes sense. So the two sides $GH \cong JK$ and the other sides $GI$ and $JL$ with $JL > GI$, so the included angles $\angle G$ (between $GH$ and $GI$) and $\angle J$ (between $JK$ and $JL$): since $JL > GI$, then $\angle J > \angle G$. Wait, but the options are $m\angle J$, $m\angle H$, $m\angle G$, $m\angle K$. So the first dropdown (left) should be $m\angle J$ and the second (right) $m\angle G$, or vice versa? Wait, the Converse Hinge Theorem says that if two sides are congruent, and the third side is longer, then the included angle is larger. So since $JL > GI$, the included angle for $JL$ (which is $\angle J$) is larger than the included angle for $GI$ (which is $\angle G$). So $m\angle J > m\angle G$.

So the correct match is $m\angle J > m\angle G$. So the first dropdown (left) is $a. m\angle J$ and the second (right) is $c. m\angle G$.

# Answer:
Left dropdown: a. $m\angle J$, Right dropdown: c. $m\angle G$ (so $m\angle J > m\angle G$)

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QUESTION IMAGE

Question

Explanation:

Step1: Recall Hinge Theorem Converse

Step2: Identify Congruent Sides and Included Angles

Answer: