Question

if $overleftrightarrow{ae} perp overleftrightarrow{ef}$ and $overleftrightarrow{bf} perp overleftrightarrow{ef}$, then $overleftrightarrow{ae}$ ? $overleftrightarrow{bf}$. options: neither, they are skew lines; $\parallel$; $\perp$

Explanation:

Step1: Recall the theorem about perpendicular lines to the same line

In a plane (or in 3D space, for lines that are coplanar or have a common reference line), if two lines are both perpendicular to the same line, then those two lines are parallel to each other. Here, \(\overleftrightarrow{AE}\perp\overleftrightarrow{EF}\) and \(\overleftrightarrow{BF}\perp\overleftrightarrow{EF}\), and from the diagram (a rectangular prism - like figure), \(AE\) and \(BF\) are in the same plane (the face containing \(A\), \(B\), \(E\), \(F\)) and both are perpendicular to \(EF\). So by the theorem, they should be parallel.

Step2: Match with the options

The second option is the symbol for parallel lines (\(\parallel\)), so that's the correct relationship between \(\overleftrightarrow{AE}\) and \(\overleftrightarrow{BF}\).

Answer:

\(\boldsymbol{\parallel}\) (the option with the two parallel lines symbol)