Question

if $overleftrightarrow{ac}perpoverleftrightarrow{ch}$ and $overleftrightarrow{ch}perpoverleftrightarrow{he}$, then $overleftrightarrow{ac}$ ? $overleftrightarrow{he}$.

Explanation:

Step1: Analyze the geometric figure

The figure is a rectangular prism (cuboid). In a cuboid, edges and lines have specific relationships. \(\overleftrightarrow{AC}\), \(\overleftrightarrow{CH}\), and \(\overleftrightarrow{HE}\) are edges (or lines related to the cuboid's structure).

Step2: Recall properties of perpendicular and parallel lines

We know that if a line is perpendicular to two other lines, and those two lines are in a plane (or form a right angle with the first line in a way that creates parallelism), then the two lines (to which the first is perpendicular) are parallel. Here, \(\overleftrightarrow{AC} \perp \overleftrightarrow{CH}\) and \(\overleftrightarrow{CH} \perp \overleftrightarrow{HE}\), and \(\overleftrightarrow{AC}\) and \(\overleftrightarrow{HE}\) are in the same direction (since they are both perpendicular to \(\overleftrightarrow{CH}\) and in the context of the cuboid, they are parallel). Also, skew lines are non - parallel and non - intersecting lines that are not in the same plane. But \(\overleftrightarrow{AC}\) and \(\overleftrightarrow{HE}\) are in parallel planes (or can be shown to be parallel) and do not intersect, but more importantly, from the perpendicularity to the same line (\(\overleftrightarrow{CH}\)), we can conclude they are parallel. So \(\overleftrightarrow{AC}\parallel\overleftrightarrow{HE}\).

Answer:

\(\parallel\) (the third option: \(\boldsymbol{\parallel}\))