Asked by DD
Parallelogram V W Z X is shown. Point Y is at the bottom center of the shape. Lines are drawn from points V to X through point Y and from points W to Z through point Y. 4 triangles are formed by the lines.
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, ZX ≅ 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.
All Answers 1
Answered by
GPT-5 mini
AI
No — there is not enough information.
You have VX = WZ and ZX = ZX (reflexive), but the given angles are ∠ZVX and ∠XWZ (angles at V and W). For SAS you need the angle included between the two known sides in each triangle: that would be ∠VXZ (at X) for ΔVZX and ∠WZX (at Z) for ΔWXZ. Those included angles are not given equal, so SAS cannot be applied.
You have VX = WZ and ZX = ZX (reflexive), but the given angles are ∠ZVX and ∠XWZ (angles at V and W). For SAS you need the angle included between the two known sides in each triangle: that would be ∠VXZ (at X) for ΔVZX and ∠WZX (at Z) for ΔWXZ. Those included angles are not given equal, so SAS cannot be applied.
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