select all that can be used to prove that the two triangles are congruent.
hypotenuse-leg
side-angle-side
angle-side-angle
angle-angle-side
angle-angle-angle

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Accepted Answer

# Brief Explanations:
1. **Hypotenuse - Leg (HL)**: HL is used for right - angled triangles only. There is no indication that these triangles are right - angled, so HL is not applicable.
2. **Side - Angle - Side (SAS)**: SAS requires two sides and the included angle. In the given triangles, we have two angles and a side, not two sides and the included angle, so SAS is not applicable.
3. **Angle - Side - Angle (ASA)**: ASA requires two angles and the included side. Let's check the angles and sides. In the first triangle, we have angles $32^{\circ}$ and $59^{\circ}$, and the side between them is $y$. In the second triangle, we have angles $32^{\circ}$ and $59^{\circ}$, and the side between them is $y$ (by correspondence). Also, we can consider the side $x$: in the first triangle, angle $32^{\circ}$, side $x$, and angle $59^{\circ}$ (the angle opposite to $y$ can be calculated as $180-(32 + 59)=89^{\circ}$, but more directly, for ASA, if we take angle $32^{\circ}$, side $x$, and angle $59^{\circ}$ (the angle adjacent to $x$ in a way that the side $x$ is between the two angles). Wait, actually, looking at the triangles, we have two angles and a side. For ASA, if we match the angles: $32^{\circ}$, side $x$, and $59^{\circ}$ (the order of angles and the included side). Also, for AAS (Angle - Angle - Side), we have two angles and a non - included side. Let's re - evaluate:
   - The sum of angles in a triangle is $180^{\circ}$. In each triangle, the third angle is $180-(32 + 59)=89^{\circ}$.
   - For ASA: Let's consider the first triangle with angles $32^{\circ}$, $59^{\circ}$ and included side $y$, and the second triangle with angles $32^{\circ}$, $59^{\circ}$ and included side $y$ (by the way the sides and angles are labeled). Also, for the side $x$: in the first triangle, angle $32^{\circ}$, side $x$, and angle $89^{\circ}$ (the third angle), and in the second triangle, angle $32^{\circ}$, side $x$, and angle $89^{\circ}$. But more precisely, for ASA, if we take angle $32^{\circ}$, side $x$, and angle $59^{\circ}$ (since $32^{\circ}$ and $59^{\circ}$ are two angles and $x$ is a side that can be considered as the included side or not? Wait, maybe a better way:
     - In the two triangles, we have:
       - Triangle 1: angles $32^{\circ}$, $59^{\circ}$, side $x$ (opposite to $59^{\circ}$? No, let's look at the labels. In the first triangle, side $x$ is between $32^{\circ}$ and the third angle, and side $y$ is between $32^{\circ}$ and $59^{\circ}$. In the second triangle, side $x$ is between $32^{\circ}$ and the third angle, and side $y$ is between $32^{\circ}$ and $59^{\circ}$.
       - For ASA: If we have two angles and the included side. Let's take angle $32^{\circ}$, side $y$, and angle $59^{\circ}$ in both triangles. So ASA is applicable.
       - For AAS (Angle - Angle - Side): We have two angles ($32^{\circ}$ and $59^{\circ}$) and a non - included side (for example, side $x$ is a non - included side with respect to the two angles $32^{\circ}$ and $59^{\circ}$ in the sense that it is opposite to one of the angles or adjacent in a non - included way). Since we know two angles and a side (either included or non - included), AAS is also applicable.
       - Angle - Angle - Angle (AAA) only proves similarity, not congruence, because it doesn't account for the side lengths. So AAA is out.
       - SAS: We don't have two sides and the included angle, so SAS is out.
       - HL: Not right triangles, so HL is out.
     - So ASA (Angle - Side - Angle) and AAS (Angle - Angle - Side) are applicable. Also, let's re - check ASA: In the first triangle, angle $32^{\circ}$, side $y$, angle $59^{\circ}$; in the second triangle, angle $32^{\circ}$, side $y$, angle $59^{\circ}$. So the included side between the two angles is $y$, so ASA works. For AAS: We have two angles ($32^{\circ}$ and $59^{\circ}$) and a side $x$ (which is a non - included side with respect to these two angles, but since the triangles have the same angle measures and a corresponding side equal, AAS also works.
# Answer:
- Angle - Side - Angle
- Angle - Angle - Side

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