Chapter 4.3, Problem 42HP

To determine

To explain:

Proof of polygon $ABED\cong FEBC$ .

Answer to Problem 42HP

Polygon $ABED\cong FEBC$ because all corresponding parts are congruent.

Explanation of Solution

Given information:

Polygon $ABED\cong FEBC$

It is given that $ABED\cong FEBC$ ,

So that $\overline{\overline{AB}}\cong \overline{\overline{EF},}\overline{BC}\cong \overline{ED},\overline{AD}\cong \overline{FC}$

Reflexive property of congruence shows that $\overline{BE}\cong \overline{EB}$

Thus, four pairs of sides are congruent.

  $\angle A\cong \angle F$ and $\angle D\cong \angle C$ , because right angles are congruent.

Use the theorem,

“Two lines in a plane that are perpendicular to the same lines are parallel”

This theorem shows $\overline{AC}\Vert \overline{DE}$ as they are perpendicular to $\overline{AD}$ .

Thus, $\angle ABE\cong \angle BEF$ and $\angle CBE\cong \angle DEB$ , because alternate interior angles are congruent.

Thus, four pairs of angles are congruent.

Hence, polygon $ABED\cong FEBC$ because all corresponding parts are congruent.