Question

question
given: (abcd) is a parallelogram and (b) is the midpoint of (overline{af}).
prove: (overline{df}) bisects (overline{bc}).
step 1: (abcd) is a parallelogram; (b) is the midpoint of (overline{af}) (reason: given)
step 2: (overline{ab} cong overline{bf}) (reason: a midpoint divides a segment into two congruent segments)
step 3: (overline{ab} cong overline{dc}) (reason: opposite sides of a parallelogram / rhombus / rectangle / square are congruent)
step 4: (overline{bf} cong overline{dc}) (reason: select a reason...)
image of parallelogram (abcd) with (f) and (e) as shown

Explanation:

Step 1: Identify the property used

We know from Step 2 that $\overline{AB} \cong \overline{BF}$ and from Step 3 that $\overline{AB} \cong \overline{DC}$. To get $\overline{BF} \cong \overline{DC}$, we use the Transitive Property of Congruence, which states that if $a \cong b$ and $b \cong c$, then $a \cong c$. Here, $a = \overline{BF}$, $b = \overline{AB}$, and $c = \overline{DC}$. So since $\overline{AB} \cong \overline{BF}$ (Step 2) and $\overline{AB} \cong \overline{DC}$ (Step 3), by the Transitive Property of Congruence, we can conclude $\overline{BF} \cong \overline{DC}$.

Step 2: State the reason for Step 4

The reason for $\overline{BF} \cong \overline{DC}$ is the Transitive Property of Congruence (if two segments are congruent to the same segment, they are congruent to each other).

Answer:

The reason for Step 4 ($\overline{BF} \cong \overline{DC}$) is the Transitive Property of Congruence (or "If two segments are congruent to the same segment, then they are congruent to each other").