Question

y is the midpoint of \\(\overline{wx}\\) and \\(\overline{wx} \perp \overline{vy}\\). complete the proof that \\(\triangle vxy \cong \triangle vwy\\).

statement	reason
2 \\(\overline{wx} \perp \overline{vy}\\)	given
3 \\(\angle vwy \cong \angle vyx\\)
4 \\(\overline{wy} \cong \overline{xy}\\)	definition of midpoint
5 \\(\overline{vy} \cong \overline{vy}\\)	reflexive property of congruence
6 \\(\triangle vxy \cong \triangle vwy\\)	sas

Explanation:

Step1: Use the definition of midpoint

Since \(Y\) is the midpoint of \(\overline{WX}\), by the definition of midpoint, \(WY = XY\), so \(\overline{WY}\cong\overline{XY}\).

Step2: Use the property of perpendicular lines

Given \(WX\perp VY\), so \(\angle VYW=\angle VYX = 90^{\circ}\), then \(\angle VYW\cong\angle VYX\).

Step3: Use the reflexive property

\(\overline{VY}\cong\overline{VY}\) by the reflexive property of congruence.

Step4: Use the SAS (Side - Angle - Side) criterion

In \(\triangle VXY\) and \(\triangle VWY\), we have \(\overline{XY}\cong\overline{WY}\), \(\angle VYX\cong\angle VYW\), \(\overline{VY}\cong\overline{VY}\). By the SAS (Side - Angle - Side) congruence criterion, \(\triangle VXY\cong\triangle VWY\).

Answer:

\(\triangle VXY\cong\triangle VWY\) by the SAS (Side - Angle - Side) congruence criterion.