Question

given: abcd is a parallelogram, e is the midpoint of \\(\overline{ab}\\) and f is the midpoint of \\(\overline{dc}\\).
prove: \\(\overline{de} \cong \overline{fb}\\).

step\tstatement\treason
1\tabcd is a parallelogram
e is the midpoint of \\(\overline{ab}\\)
f is the midpoint of \\(\overline{dc}\\)\tgiven
2\t\\(\overline{ad} \cong \overline{cb}\\)\topposite sides of a parallelogram / rhombus / rectangle / square are congruent
3\t\\(\overline{dc} \cong \overline{ab}\\)\topposite sides of a parallelogram / rhombus / rectangle / square are congruent
4\t\\(fc = \frac{1}{2}dc\\)\twhen a midpoint divides a segment into two parts, each part is half the length of the whole segment
5\t\\(ae = \frac{1}{2}ab\\)\tselect a reason...

Explanation:

Step1: Analyze the midpoint property

Since \( E \) is the midpoint of \( \overline{AB} \), by the definition of a midpoint (a midpoint divides a segment into two congruent segments, each half the length of the whole segment), we have \( AE=\frac{1}{2}AB \). The reason is the same as step 4, where for midpoint \( F \) of \( \overline{DC} \), we used the property that a midpoint divides a segment into two parts, each part is half the length of the whole segment. So the reason for \( AE = \frac{1}{2}AB \) is "When a midpoint divides a segment into two parts, each part is half the length of the whole segment".

To continue proving \( \overline{DE}\cong\overline{FB} \):

Step6: Use midpoint and parallelogram properties

From step 3, \( \overline{DC}\cong\overline{AB} \), so \( DC = AB \) (by definition of congruent segments). From step 4, \( FC=\frac{1}{2}DC \) and from step 5 (now with reason), \( AE=\frac{1}{2}AB \). Since \( DC = AB \), then \( FC = AE \). Also, from step 2, \( \overline{AD}\cong\overline{CB} \), so \( AD = CB \). And since \( ABCD \) is a parallelogram, \( \overline{AB}\parallel\overline{DC} \), so \( \overline{AE}\parallel\overline{FC} \) (as \( E \) is on \( AB \) and \( F \) is on \( DC \)). So quadrilateral \( AECF \) is a parallelogram? Wait, no, we need to look at quadrilateral \( DEBF \).

Wait, \( AB = DC \), \( E \) and \( F \) are midpoints, so \( EB=\frac{1}{2}AB \) and \( DF=\frac{1}{2}DC \), so \( EB = DF \). Also, \( AB\parallel DC \), so \( EB\parallel DF \) (since \( E \) is on \( AB \) and \( F \) is on \( DC \)). So quadrilateral \( DEBF \) has \( EB\parallel DF \) and \( EB = DF \), so it is a parallelogram (by definition of a parallelogram: a quadrilateral with one pair of opposite sides parallel and congruent is a parallelogram). In a parallelogram, opposite sides are congruent, so \( \overline{DE}\cong\overline{FB} \).

But let's structure the steps properly:

Step5: Reason for \( AE=\frac{1}{2}AB \)

Statement: \( AE = \frac{1}{2}AB \)
Reason: When a midpoint divides a segment into two parts, each part is half the length of the whole segment (since \( E \) is the midpoint of \( \overline{AB} \))

Step6: Find \( EB \) and \( DF \) lengths

Since \( E \) is the midpoint of \( AB \), \( EB=\frac{1}{2}AB \). Since \( F \) is the midpoint of \( DC \), \( DF=\frac{1}{2}DC \). From step 3, \( AB = DC \), so \( EB = DF \).

Step7: Parallel sides

Since \( ABCD \) is a parallelogram, \( AB\parallel DC \), so \( EB\parallel DF \) (as \( EB \) is part of \( AB \) and \( DF \) is part of \( DC \)).

Step8: Conclude \( DEBF \) is parallelogram

A quadrilateral with one pair of opposite sides parallel and congruent is a parallelogram. So \( DEBF \) is a parallelogram (since \( EB\parallel DF \) and \( EB = DF \)).

Step9: Opposite sides of parallelogram

In parallelogram \( DEBF \), opposite sides are congruent, so \( \overline{DE}\cong\overline{FB} \).

Answer:

For step 5, the reason is "When a midpoint divides a segment into two parts, each part is half the length of the whole segment". And the final proof shows \( \overline{DE}\cong\overline{FB} \) by proving \( DEBF \) is a parallelogram (using midpoint properties, parallelogram side congruence and parallelism) and then using the property that opposite sides of a parallelogram are congruent.