if vx = wz = 40 cm and m∠zvx = m∠xwz = 22°, can δvzx and δwxz be proven congruent by sas? why or why not? yes, along with the given information, $ \overline{zx} \cong \overline{zx} $ by the reflexive property. yes, the triangles are both obtuse. no, the sides of the triangles intersect. no, there is not enough information given.

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# Brief Explanations:
To prove two triangles congruent by SAS (Side - Angle - Side), we need two sides and the included angle of one triangle to be congruent to the corresponding two sides and included angle of the other triangle.

In $\triangle VZX$ and $\triangle WXZ$, we are given $VX = WZ=40\mathrm{cm}$ and $\angle ZVX=\angle XWZ = 22^{\circ}$. The side $ZX$ is common to both triangles ($\overline{ZX}\cong\overline{ZX}$ by the reflexive property of congruence).

The angle $\angle ZVX$ is between sides $VX$ and $ZV$ in $\triangle VZX$, and the angle $\angle XWZ$ is between sides $WZ$ and $WX$ in $\triangle WXZ$. But we don't know if $ZV = WX$.

However, for SAS, the angle must be the included angle between the two sides. Here, the sides $VX$ and $ZX$ with included angle $\angle VXZ$ in $\triangle VZX$ and  sides $WZ$ and $ZX$ with included angle $\angle WZX$ in $\triangle WXZ$ are not considered. But when we look at the given information $VX = WZ$, $\angle ZVX=\angle XWZ$ and $ZX = ZX$ (reflexive property), the sides - angle - sides (the side $VX$ in $\triangle VZX$, the angle $\angle ZVX$, the side $ZX$ in $\triangle VZX$ and the side $WZ$ in $\triangle WXZ$, the angle $\angle XWZ$, the side $ZX$ in $\triangle WXZ$)  satisfy the SAS congruence criterion.

# Answer:
Yes, along with the given information, $\overline{ZX}\cong\overline{ZX}$ by the reflexive property.

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