Question

which congruence theorem can be used to prove △wxs ≅ △yzs? diagram of triangles with right angles and marked segments options: sss, asa, sas, hl

Explanation:

Step1: Analyze given angles and sides

From the diagram, $\angle WXS$ and $\angle YZS$ are right angles (so $\angle WXS = \angle YZS = 90^\circ$). Also, $XS = ZS$ (marked with equal segments), and $\angle WSX$ and $\angle YSZ$ are vertical angles, so $\angle WSX=\angle YSZ$.

Step2: Identify congruence theorem

We have two angles and the included side? Wait, no: Wait, $\angle WXS = \angle YZS$ (right angles), $XS = ZS$ (side), and $\angle WSX=\angle YSZ$ (vertical angles). Wait, actually, let's check: In $\triangle WXS$ and $\triangle YZS$:

• $\angle WXS = \angle YZS$ (both right angles, so $90^\circ$)
• $XS = ZS$ (given by the marks)
• $\angle WSX = \angle YSZ$ (vertical angles are congruent)

So this is Angle - Side - Angle (ASA) because we have two angles and the included side? Wait, no: Wait, the side is between the two angles? Wait, in $\triangle WXS$, the angles at $X$ and $S$, with side $XS$ between them. In $\triangle YZS$, angles at $Z$ and $S$, with side $ZS$ between them. Since $XS = ZS$, $\angle WXS=\angle YZS$, $\angle WSX=\angle YSZ$, so by ASA (Angle - Side - Angle) congruence theorem, the triangles are congruent. Wait, but also, let's check other options: SSS would require three sides, we don't have that. SAS would require two sides and included angle, but here we have two angles and a side. HL is for right triangles with hypotenuse and leg, but we don't know hypotenuse yet. So ASA is correct? Wait, no, wait: Wait, the right angles, the vertical angles, and the side between them? Wait, maybe I made a mistake. Wait, alternatively, let's see: $\angle WXS$ and $\angle YZS$ are right angles (so right triangles), $XS = ZS$ (leg), and $\angle WSX=\angle YSZ$ (angle). Wait, but ASA: two angles and the included side. So angle at $X$, side $XS$, angle at $S$: yes, that's ASA. Wait, but let's confirm the options. The options are SSS, ASA, SAS, HL. So ASA is the correct theorem here because we have two angles (right angle and vertical angle) and the included side (XS = ZS) between them.

Wait, no, wait: Wait, in $\triangle WXS$ and $\triangle YZS$:

• $\angle WXS = \angle YZS$ (right angles, so congruent)
• $XS = ZS$ (given, side)
• $\angle WSX = \angle YSZ$ (vertical angles, congruent)

So this is Angle - Side - Angle (ASA) because the side is between the two angles. So the congruence theorem is ASA.

Answer:

ASA (the option corresponding to ASA, e.g., if the options are labeled as above, the answer is the option with ASA, like "ASA" or the option letter if there was a letter, but in the given options, it's the ASA option)