Question

which rule explains why these triangles are congruent?
sss
asa
aas
sas
these triangles cannot be proven congruent.

Explanation:

1. Identify corresponding parts: In $\triangle FJG$ and $\triangle HGF$, $FJ = HG$ (marked congruent), $\angle J = \angle G$ (marked right angles), and $FG = HF$ is not correct—instead, the shared side is $FH$ for $\triangle FJH$ and $FG$ for $\triangle HGF$? No, correct pairs: $FJ = HG$ (marked), $\angle J = \angle G$ (right angles), and $FH$ is a common side? No, $\triangle FJH$ and $\triangle HGF$: $FJ = HG$, $\angle J = \angle G$, $JH = FG$? No, the marked sides are $FJ$ and $HG$, right angles at $J$ and $G$, and side $FH$ is common? No, $FH$ is a side of both triangles. Wait, $\triangle FJH$: sides $FJ, JH, FH$; $\triangle HGF$: sides $HG, GF, FH$. $FJ=HG$, $\angle J=\angle G$ (right angles), $FH=FH$? No, that would be SSA, which is not valid. Wait no, the marked sides are $FJ$ and $HG$, right angles at $J$ and $G$, and $JH$ and $FG$ are not marked. Wait, no: $\triangle FJH$ and $\triangle HGF$: $\angle J = \angle G$ (right angles), $FJ = HG$ (marked), and $\angle JFH = \angle GHF$? No, alternate interior angles? No, the figure is a quadrilateral with $FJ \parallel HG$? Because $\angle J$ and $\angle G$ are right angles, so $FJ \parallel HG$, and $FJ=HG$, so it's a parallelogram, so $JH=FG$, and $FH$ is diagonal. So $\triangle FJH$ and $\triangle HGF$: $FJ=HG$, $JH=FG$, $FH=HF$: that's SSS? No, wait the marked sides are only $FJ$ and $HG$, right angles at $J$ and $G$. Wait, the correct rule here is AAS: $\angle J = \angle G$, $\angle JFH = \angle GHF$ (alternate interior angles, since $FJ \parallel HG$), and $FJ=HG$. That's two angles and a non-included side, which is AAS. Wait no, the marked parts are $\angle J = \angle G$, $FJ=HG$, and $FH$ is common? No, no, the correct way: in $\triangle FJH$ and $\triangle HGF$:

• $\angle J = \angle G$ (right angles, marked)
• $FJ = HG$ (marked congruent)
• $\angle JHF = \angle GFH$ (alternate interior angles, as $FJ \parallel HG$ cut by transversal $FH$)

This is two angles and the included side? No, $FJ$ is between $\angle J$ and $\angle JFH$, $HG$ is between $\angle G$ and $\angle GHF$. So $\angle J = \angle G$, $FJ=HG$, $\angle JFH = \angle GHF$: this is ASA? No, wait no, the side is between the two angles. Wait, no, the marked parts are only one side and one angle, plus the common side? No, no, the figure shows:

• $\angle J$ and $\angle G$ are congruent (right angles)
• $FJ$ and $HG$ are congruent (marked with one tick)
• $FH$ is a common side to both triangles $\triangle FJH$ and $\triangle HGF$

Wait, that would be SSA, which is not a valid congruence rule. But wait, no, the triangles are $\triangle FJH$ and $\triangle HGF$: $FJ=HG$, $\angle J=\angle G$, $JH=FG$? No, $JH$ and $FG$ are not marked. Wait, no, the correct congruence rule here is AAS: we have two angles and a non-included side. $\angle J = \angle G$, $\angle JFH = \angle GHF$, and $FJ=HG$. That's AAS.

Answer:

AAS