Question

which congruence theorem can be used to prove △wxz ≅ △yzx? diagram of triangles with vertices w, x, z, y (right angles at x and z, angles marked at w and y) options: aas, asa, sas, hl

Explanation:

1. First, identify the right angles: $\angle WXZ$ and $\angle YZX$ are right angles (so $\angle WXZ = \angle YZX = 90^\circ$).
2. The side $XZ$ is common to both $\triangle WXZ$ and $\triangle YZX$, so $XZ = ZX$ (reflexive property).
3. We also have $\angle W$ and $\angle Y$ as non - right angles, and we can see that we have two angles and a non - included side? Wait, no. Wait, for AAS: two angles and a non - included side. But let's re - examine. Wait, actually, in right triangles, the HL (Hypotenuse - Leg) theorem states that if the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and one leg of another right triangle, then the triangles are congruent. Wait, no, let's check the angles and sides again. Wait, $\triangle WXZ$ and $\triangle YZX$ are right triangles (right - angled at $X$ and $Z$ respectively). The hypotenuse $WZ$ and $YX$? Wait, no, maybe I made a mistake. Wait, the common side is $XZ$, and we have $\angle W=\angle Y$ (from the diagram, the marked angles), $\angle WXZ=\angle YZX = 90^\circ$, and side $XZ = ZX$. So we have two angles and a non - included side? Wait, no, AAS is two angles and a non - included side. But wait, actually, let's check the options. Wait, HL is for right triangles: hypotenuse and leg. But in this case, let's see: $\triangle WXZ$ and $\triangle YZX$ are right triangles. The leg $XZ$ is common (so $XZ = ZX$), and we have another pair of angles equal ($\angle W=\angle Y$) and the right angles. Wait, no, maybe it's AAS? Wait, no, let's think again. Wait, the correct approach:

• $\angle WXZ=\angle YZX = 90^\circ$ (right angles).
• $\angle W=\angle Y$ (given by the marked angles in the diagram).
• $XZ = ZX$ (common side, reflexive property).

So we have two angles and a non - included side (AAS: Angle - Angle - Side). Wait, but wait, AAS is when two angles and a non - included side are congruent. In this case, the two angles are $\angle W$ and $\angle WXZ$ in $\triangle WXZ$, and $\angle Y$ and $\angle YZX$ in $\triangle YZX$, and the side $XZ$ (which is opposite to $\angle W$ in $\triangle WXZ$ and opposite to $\angle Y$ in $\triangle YZX$). So AAS congruence theorem applies. Wait, but let's check the other options:

• ASA: Angle - Side - Angle, which requires the side to be included between the two angles. Here, the side $XZ$ is not included between $\angle W$ and $\angle WXZ$ (the included side would be $WX$), so ASA is not correct.
• SAS: Side - Angle - Side, we don't have two sides and the included angle.
• HL: Hypotenuse - Leg, we don't know about the hypotenuse being congruent yet, and we have two angles equal, so HL is not the case.

So the correct congruence theorem is AAS.

Answer:

A. AAS