Question

complete the proof that $\triangle rsw \cong \triangle utv$.

statement	reason
2 $\angle tvu \cong \angle rws$	given
3 $\overline{st} \cong \overline{vw}$	given
4 $sw = vw + sv$	additive property of length
5 $tv = st + sv$	additive property of length
6 $sw = st + sv$	substitution
7

Explanation:

Step1: Analyze Step 5 and 6

From step 5, \( TV = ST + SV \), and step 6 gives \( SW = ST + SV \). By the Transitive Property of Equality, if \( a = b \) and \( b = c \), then \( a = c \). So \( SW = TV \).

Step2: Identify Congruent Triangles

We have two angles (\( \angle RSW \cong \angle UTV \), \( \angle TVU \cong \angle RWS \)) and the included side (\( SW \cong TV \)) congruent. By the AAS (Angle - Angle - Side) Congruence Criterion, \( \triangle RSW \cong \triangle UTV \). But for the statement in step 7, since we showed \( SW = TV \) (from step 5 and 6), the statement should be \( \overline{SW} \cong \overline{TV} \) (or \( SW = TV \)) and the reason is Transitive Property (or Substitution from step 5 and 6). But to complete the proof for congruence, after having two angles and a side, the side we just proved equal is the non - included side? Wait, no, let's re - check. Wait, \( \angle RSW \) and \( \angle UTV \) are angles, \( \angle TVU \) and \( \angle RWS \) are angles, and the side between? Wait, no, \( \angle RSW \) is at \( S \), \( \angle RWS \) is at \( W \), so the side between them is \( SW \). For \( \triangle UTV \), \( \angle UTV \) is at \( T \), \( \angle TVU \) is at \( V \), so the side between them is \( TV \). Since we proved \( SW = TV \), and we have two angles congruent, by AAS, the triangles are congruent. But the statement in step 7 should be \( SW = TV \) (or \( \overline{SW} \cong \overline{TV} \)) with reason Transitive Property (because \( TV = ST + SV \) and \( SW = ST + SV \), so \( SW = TV \)).

Step3: Write the Statement for Step 7

The statement for step 7 is \( SW = TV \) (or \( \overline{SW} \cong \overline{TV} \)) and the reason is Transitive Property of Equality (or Substitution, since \( TV = ST + SV \) and \( SW = ST + SV \)). But to match the proof structure, since we are building up to AAS, the side we need to show congruent is \( SW \) and \( TV \). So the statement is \( \overline{SW} \cong \overline{TV} \) (or \( SW = TV \)) and the reason is Transitive Property (or Substitution from step 5 and 6).

Answer:

The statement for step 7 is \( \boldsymbol{SW = TV} \) (or \( \boldsymbol{\overline{SW} \cong \overline{TV}} \)) and the reason is Transitive Property of Equality (or Substitution). (If we follow the proof steps, the most direct statement after step 5 and 6 is \( SW = TV \) with reason Transitive Property because \( TV = ST + SV \) and \( SW = ST + SV \), so \( SW = TV \).)