QUESTION IMAGE
Question
1 z is the midpoint of \\( \overline { w y } \\) given
2 z is the midpoint of \\( \overline { v x } \\) given
3 \\( \overline { w z } \cong \overline { y z } \\)
4 \\( \overline { v z } \cong \overline { x z } \\) definition of midpoint
5 \\( \angle v z y \cong \angle w z x \\) vertical angle theorem
6 \\( \triangle v y z \cong \triangle x w z \\)
Step1: Recall the definition of mid - point
If \(Z\) is the mid - point of a line segment \(\overline{AB}\), then \(AZ = BZ\) (by the definition of mid - point). Since \(Z\) is the mid - point of \(\overline{WY}\), we have \(\overline{WZ}\cong\overline{YZ}\) (by the definition of mid - point).
Step2: Use the Side - Angle - Side (SAS) congruence criterion
The SAS congruence criterion states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the two triangles are congruent.
We know that \(\overline{WZ}\cong\overline{YZ}\), \(\overline{VZ}\cong\overline{XZ}\) (from step 1 and given mid - point for \(VX\)), and \(\angle VZY\cong\angle WZX\) (vertical angles).
For \(\triangle VYZ\) and \(\triangle XWZ\), in \(\triangle VYZ\) and \(\triangle XWZ\):
- \( \overline{WZ}=\overline{YZ}\) (from step 1)
- \( \overline{VZ}=\overline{XZ}\) (given mid - point of \(VX\))
- \( \angle VZY=\angle WZX\) (vertical angles)
By the SAS (Side - Angle - Side) congruence criterion, \(\triangle VYZ\cong\triangle XWZ\)
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- Definition of mid - point; 6. SAS (Side - Angle - Side) congruence criterion.