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 | statement | reason
1 | abcd is a parallelogram; e is the midpoint of \\(\overline{ab}\\); f is the midpoint of \\(\overline{dc}\\) | given
2 | \\(\overline{ad} \cong \overline{cb}\\) | opposite sides of a parallelogram / rhombus / rectangle / square are congruent
3 | \\(\overline{dc} \cong \overline{ab}\\) | opposite sides of a parallelogram / rhombus / rectangle / square are congruent
4 | \\(fc = \frac{1}{2}dc\\) | select a reason...

Explanation:

Step 1: Identify the reason for \( FC = \frac{1}{2}DC \)

Since \( F \) is the midpoint of \( \overline{DC} \), by the definition of a midpoint, a midpoint divides a segment into two congruent segments, each equal to half the length of the original segment. So the reason is "Definition of a midpoint (a midpoint divides a segment into two equal parts)".

Step 2: Find \( EB \) in terms of \( AB \)

Similarly, \( E \) is the midpoint of \( \overline{AB} \), so \( EB=\frac{1}{2}AB \) (by the definition of a midpoint).

Step 3: Use the property of parallelogram to relate \( DC \) and \( AB \)

From step 3, we know \( \overline{DC}\cong\overline{AB} \), which implies \( DC = AB \) (congruent segments have equal length).

Step 4: Show \( FC = EB \)

Since \( FC=\frac{1}{2}DC \) and \( EB = \frac{1}{2}AB \), and \( DC = AB \), by substitution, \( FC=EB \).

Step 5: Prove \( \triangle ADE \cong \triangle CBF \) (or \( \triangle DFC \cong \triangle EBA \)) or prove \( DEBF \) is a parallelogram

We already have \( \overline{AD}\cong\overline{CB} \) (step 2), \( \angle DAB\cong\angle BCD \) (opposite angles of a parallelogram are congruent), and \( EB = FC \) (step 4). Wait, alternatively, since \( AB\parallel DC \) (opposite sides of a parallelogram are parallel) and \( EB = FC \) and \( EB\parallel FC \) (because \( AB\parallel DC \) and \( E,F \) are midpoints), so quadrilateral \( DEBF \) has one pair of opposite sides equal and parallel, hence it is a parallelogram. In a parallelogram, opposite sides are congruent, so \( \overline{DE}\cong\overline{FB} \).

But focusing on step 4, the reason for \( FC=\frac{1}{2}DC \) is the definition of a midpoint.

Answer:

For step 4, the reason is "Definition of a midpoint (a midpoint divides a segment into two equal - length segments, so each segment is half the length of the original segment)". And to complete the proof:

1. From step 4: \( FC=\frac{1}{2}DC \) (Definition of midpoint).
2. Since \( E \) is the midpoint of \( \overline{AB} \), \( EB = \frac{1}{2}AB \) (Definition of midpoint).
3. From step 3, \( DC = AB \) (because \( \overline{DC}\cong\overline{AB} \), congruent segments have equal length).
4. So \( FC=EB \) (by substitution, since \( FC = \frac{1}{2}DC \), \( EB=\frac{1}{2}AB \) and \( DC = AB \)).
5. Also, \( AB\parallel DC \) (opposite sides of a parallelogram are parallel), so \( EB\parallel FC \) (since \( EB \) is part of \( AB \) and \( FC \) is part of \( DC \)).
6. A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram (so \( DEBF \) is a parallelogram).
7. In a parallelogram, opposite sides are congruent, so \( \overline{DE}\cong\overline{FB} \).

And the reason for step 4 is "Definition of a midpoint".