Question

proof: complete the paragraph proof
given: v is the midpoint of \\(\overline{wx}\\) and \\(\overline{yz}\\).
prove: \\(\triangle xvz \cong \triangle wvy\\)
proof:
because v is the midpoint of \\(\overline{yz}\\) and the midpoint of \\(\overline{wx}\\), by the midpoint theorem, segment \\(yv \cong\\) segment select choice and segment \\(wv \cong\\) segment select choice. the angles are congruent. therefore, by select choice, \\(\triangle xvz \cong \triangle wvy\\).
image of two triangles sharing a common vertex v, with vertices labeled w, y, v, x, z

Explanation:

Step1: Identify midpoint segments

Since \( V \) is the midpoint of \( YZ \), by the definition of a midpoint, \( YV = ZV \) (so \( YV \cong ZV \)).
Since \( V \) is the midpoint of \( WX \), by the definition of a midpoint, \( XV = WV \) (so \( XV \cong WV \)).

Step2: Identify angle relationship

\( \angle YVW \) and \( \angle ZVX \) are vertical angles. By the Vertical Angles Theorem, vertical angles are congruent, so \( \angle YVW\cong\angle ZVX \).

Step3: Apply triangle congruence criterion

We have two pairs of congruent sides (\( YV \cong ZV \), \( XV \cong WV \)) and the included angle (\( \angle YVW\cong\angle ZVX \)) congruent. By the Side - Angle - Side (SAS) Congruence Theorem, \( \triangle XVZ\cong\triangle WVY \) (note: the original problem has a typo, it should be \( \triangle XVZ\cong\triangle WVY \) or adjusted based on the diagram, but the SAS criterion applies here with the identified sides and angle).

Answer:

1. First "Select Choice": \( YV \cong ZV \) (since \( V \) is midpoint of \( YZ \), so \( YV = ZV \))
2. Second "Select Choice": \( XV \cong WV \) (since \( V \) is midpoint of \( WX \), so \( XV = WV \))
3. Third "Select Choice": Vertical angles ( \( \angle YVW \) and \( \angle ZVX \) are vertical angles, so they are congruent)
4. Fourth "Select Choice": SAS (Side - Angle - Side, as we have two pairs of congruent sides and a pair of congruent included angles)