Question

2 construct \\(\overline{be}\\) perpendicular to \\(\overleftrightarrow{cd}\\) such that point \\(e\\) is on \\(\overleftrightarrow{cd}\\).
3 construct \\(\overline{cf}\\) perpendicular to \\(\overleftrightarrow{ab}\\) such that point \\(f\\) is on \\(\overleftrightarrow{ab}\\).
4 \\(m\angle cfb = m\angle bec = 90^\circ\\) all perpendicular angles measure \\(90^\circ\\) (2, 3).
5 \\(cf = \boldsymbol{\downarrow}\\) any point on one parallel line is the same distance from the other line on a perpendicular transversal (2, 3).
6 \\(bc = bc\\) they are measures of the same segment.
7 \\(\triangle bcf \cong \triangle cbe\\) congruence (6, 5)
8 \\(\angle fbc \cong \angle ecb\\) corresponding parts of congruent figures are congruent

Explanation:

Step1: Recall the property of parallel lines and perpendicular distance

The problem states that \( AB \) and \( CD \) are parallel lines (implied by the construction of perpendiculars from points on one line to the other). The key property here is that the distance between two parallel lines is constant, and the perpendicular segments between them are equal in length. Since \( BE \) is perpendicular to \( CD \) and \( CF \) is perpendicular to \( AB \), and \( AB \parallel CD \), the lengths of \( CF \) and \( BE \) should be equal.

Step2: Apply the property to find \( CF \)

Using the property that any point on one parallel line is the same distance from the other line along a perpendicular transversal, we conclude that \( CF = BE \).

Answer:

\( BE \)