question
given: abcd is a parallelogram and b is the midpoint of $\overline{af}$.
prove: $\overline{df}$ bisects $\overline{bc}$.
step\tstatement\treason
1\tabcd is a parallelogram
b is the midpoint of $\overline{af}$\tgiven
2\t$\overline{ab} \cong \overline{bf}$\ta midpoint divides a segment into two congruent segments
3\t$\overline{ab} \cong \overline{dc}$\tselect a reason...
image of a parallelogram abcd with b as midpoint of af, and df intersecting bc at e

Question

Accepted Answer

# Explanation:
## Step1: Recall parallelogram properties
In a parallelogram, opposite sides are congruent. Since $ABCD$ is a parallelogram, $AB$ and $DC$ are opposite sides. So, by the definition of a parallelogram (opposite sides of a parallelogram are congruent), we have $\overline{AB} \cong \overline{DC}$.
## Step2: Use the midpoint property (already used for $AB \cong BF$) and the parallelogram side congruence
We know from step 2 that $\overline{AB} \cong \overline{BF}$ and from step 3 (our new step) that $\overline{AB} \cong \overline{DC}$. By the transitive property of congruence (if $a \cong b$ and $b \cong c$, then $a \cong c$), we can say $\overline{BF} \cong \overline{DC}$.
## Step3: Identify parallel lines and alternate interior angles
Since $ABCD$ is a parallelogram, $AB \parallel DC$. Also, $AB$ and $BF$ are on the same line $AF$, so $DC \parallel BF$. Now, consider lines $DF$ (the transversal) and $BC$ (intersected by $DC$ and $BF$). The alternate interior angles $\angle CDE$ and $\angle BFE$ are congruent, and $\angle DCE$ and $\angle FBE$ are congruent (because $DC \parallel BF$ and $DF$ is a transversal).
## Step4: Prove triangles congruent
We have $\overline{BF} \cong \overline{DC}$ (from step 2 and 3), $\angle CDE \cong \angle BFE$, and $\angle DCE \cong \angle FBE$. By the Angle - Side - Angle (ASA) congruence criterion, $\triangle DCE \cong \triangle FBE$.
## Step5: Conclude that $DF$ bisects $BC$
Since $\triangle DCE \cong \triangle FBE$, their corresponding sides $CE$ and $BE$ are congruent (corresponding parts of congruent triangles are congruent, CPCTC). So, $E$ is the midpoint of $BC$, which means $\overline{DF}$ bisects $\overline{BC}$ (a segment bisector is a line that divides the segment into two congruent parts, and since $BE = CE$, $DF$ bisects $BC$).

But for the specific step 3 in the given proof (finding the reason for $\overline{AB} \cong \overline{DC}$):
The reason for $\overline{AB} \cong \overline{DC}$ is "Opposite sides of a parallelogram are congruent".

# Answer:
The reason for $\overline{AB} \cong \overline{DC}$ is "Opposite sides of a parallelogram are congruent".

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QUESTION IMAGE

Question

Explanation:

Step1: Recall parallelogram properties

Step2: Use the midpoint property (already used for \(AB \cong BF\)) and the parallelogram side congruence

Step3: Identify parallel lines and alternate interior angles

Step4: Prove triangles congruent

Step5: Conclude that \(DF\) bisects \(BC\)

Answer: