Question

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

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 \(\overline{WX}\perp\overline{VY}\), then \(\angle VWY=\angle VXY = 90^{\circ}\), so \(\angle VWY\cong\angle VXY\)

Step3: Use the reflexive property

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

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

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

Answer:

\(\triangle VXY\cong\triangle VWY\) (by SAS congruence criterion)