Question

in $\triangle wxy$, $\overline{wz}$ is a median.

which additional condition would not be sufficient to prove that $\triangle xwz \cong \triangle ywz?$

a $\overline{xw} \cong \overline{wy}$

b $\angle x \cong \angle y$

c $\overline{xy} \cong \overline{wz}$

Explanation:

Step1: Recall median property

Since $\overline{WZ}$ is a median, $\overline{XZ} \cong \overline{ZY}$. Also, $\overline{WZ} \cong \overline{WZ}$ (reflexive property).

Step2: Analyze Option A

Condition: $\overline{XW} \cong \overline{WY}$.
We have 3 pairs of congruent sides: $\overline{XW} \cong \overline{WY}$, $\overline{XZ} \cong \overline{ZY}$, $\overline{WZ} \cong \overline{WZ}$. This satisfies SSS congruence, so $\triangle XWZ \cong \triangle YWZ$.

Step3: Analyze Option B

Condition: $\angle X \cong \angle Y$.
We have $\angle X \cong \angle Y$, $\overline{XZ} \cong \overline{ZY}$, $\overline{WZ} \cong \overline{WZ}$? No, this is SSA, which does NOT prove congruence. Wait, correction: We have $\angle X \cong \angle Y$, $\overline{XZ} \cong \overline{ZY}$, and $\angle WZX$ vs $\angle WZY$? No, actually, with $\overline{XW} \cong \overline{WY}$ implied? No, wait: $\angle X \cong \angle Y$, $\overline{XZ} \cong \overline{ZY}$, $\overline{WZ}$ is common. This is SSA, which is not valid for congruence, but wait no—wait, if $\angle X \cong \angle Y$, $\overline{XZ} \cong \overline{ZY}$, $\overline{WX}$ and $\overline{WY}$: actually, this is AAS? No, $\angle X$, $\angle WZX$, side? No, $\angle WZX$ and $\angle WZY$ are not known. Wait, no: if $\overline{XW} \cong \overline{WY}$, then $\triangle WXY$ is isosceles, so $\angle X \cong \angle Y$, and then $\triangle XWZ \cong \triangle YWZ$ by SAS ( $\overline{XZ} \cong \overline{ZY}$, $\angle X \cong \angle Y$, $\overline{XW} \cong \overline{WY}$? No, SAS is side-angle-side. Wait, $\overline{XZ} \cong \overline{ZY}$, $\angle X \cong \angle Y$, $\overline{XW} \cong \overline{WY}$ is SAS? No, the angle must be included. Oh right, $\angle X$ is between $\overline{XW}$ and $\overline{XZ}$, $\angle Y$ is between $\overline{YW}$ and $\overline{YZ}$. So yes, that is SAS congruence. So $\triangle XWZ \cong \triangle YWZ$ is proven here.

Step4: Analyze Option C

Condition: $\overline{XY} \cong \overline{WZ}$.
We know $\overline{XY} = 2\overline{XZ}$ (since Z is midpoint), so $\overline{WZ} = 2\overline{XZ}$. This gives us $\overline{XZ} \cong \overline{ZY}$, $\overline{WZ} \cong \overline{WZ}$, but $\overline{XY} \cong \overline{WZ}$ does not give us a congruent side or angle to satisfy any congruence postulate (SSS, SAS, ASA, AAS, HL). This condition does not prove $\triangle XWZ \cong \triangle YWZ$.

Answer:

C. $\overline{XY} \cong \overline{WZ}$