Questions: triangle DEF-

Solution Steps

We are given two triangles, $\triangle MNV$ and $\triangle FEZ$. We are given that $\angle MNV = 60^{\circ}$ and $\angle FEZ = 60^{\circ}$. Since $MN$ and $FE$ are parallel, we can say that $\angle NMD = \angle FED$ and $\angle MND = \angle EFD$. In $\triangle MNV$, $\angle MNV = 60^{\circ}$. Since the sum of angles in a triangle is $180^{\circ}$, we have: $\angle NMV + \angle MNV + \angle NVM = 180^{\circ}$ $\angle NMV + 60^{\circ} + \angle NVM = 180^{\circ}$ $\angle NMV + \angle NVM = 120^{\circ}$

In $\triangle FEZ$, $\angle FEZ = 60^{\circ}$. Since the sum of angles in a triangle is $180^{\circ}$, we have: $\angle EFZ + \angle FEZ + \angle EZF = 180^{\circ}$ $\angle EFZ + 60^{\circ} + \angle EZF = 180^{\circ}$ $\angle EFZ + \angle EZF = 120^{\circ}$

We are given that $\angle MNV = \angle FEZ = 60^{\circ}$. Since $MN \parallel FE$, we have $\angle NMD = \angle FED$ and $\angle MND = \angle EFD$. Since $\angle NMV + \angle NVM = 120^{\circ}$ and $\angle EFZ + \angle EZF = 120^{\circ}$, we cannot determine if $\angle NMV = \angle EFZ$ or $\angle NVM = \angle EZF$. However, if the lines $ME$ and $NF$ are parallel, then the two triangles would be similar. We can also say that if $\angle NMD = \angle FED$ and $\angle MND = \angle EFD$, then $\angle NVM = \angle EZF$ since $MN \parallel FE$. In that case $\triangle MNV \sim \triangle FEZ$ using the AA similarity criteria.

Final Answer: