Question

complete the proof that the alternate interior angles of transversals of parallel lines are congruent. note: this proof is for the case where m∠1 is less than 90°. this proof uses the following theorem: any point on one parallel line is the same distance from the other line on a perpendicular transversal. statement or construction reason 1 $overleftrightarrow{ab}paralleloverleftrightarrow{cd}$ given 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∠cfb = m∠bec = 90° all perpendicular angles measure 90° (2, 3). 5 cf = blank any point on one parallel line is the same distance from the other line on a perpendicular transversal (1, 2, 3). 6 bc = bc they are measures of the same segment.

Explanation:

Step1: Recall the given theorem

Since any point on one parallel line is the same distance from the other line on a perpendicular transversal, and we have constructed perpendiculars $CF$ and $BE$ between parallel lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{CD}$, the lengths of the perpendicular - segments are equal.

Step2: Fill in the blank

The blank in statement 5 should be filled with $BE$ because of the property that the distance between two parallel lines is the same along any perpendicular transversal. So $CF = BE$.

Answer:

$BE$