1. $\overline{ab} \cong \overline{fg}$
2. $\overline{bf}$ bisects $\overline{ac}$ and $\overline{dg}$.
3. $\overline{bc} \cong \overline{ab}$, $\overline{fg} \cong \overline{df}$
4. $\overline{bc} \cong \overline{fg}$
5. $\overline{bc} \cong \overline{df}$
left side justifications:
$\boldsymbol{\because} \overline{bc} \cong \overline{fg}$
$\boldsymbol{\because} \overline{bc} \cong \overline{ab}$, $\overline{fg} \cong \overline{df}$
$\boldsymbol{\because}$ definition of segment bisector
$\boldsymbol{\because}$ transitive property of equality
$\boldsymbol{\because}$ transitive property of segment congruence
bottom labels:
1. given
2. given
3. definition of segment bisector
4. transitive property of segment congruence
5. (blank)

Question

Accepted Answer

To solve this geometric proof, we analyze each step based on the given information and properties of segment congruence:

### Step 1: Analyze Given Information
- We know $ AB \cong FG $ (Given) and $ BF $ bisects $ AC $ and $ DG $ (Given). By the definition of a segment bisector, a bisector divides a segment into two congruent segments. So, if $ BF $ bisects $ AC $, then $ AB \cong BC $, and if $ BF $ bisects $ DG $, then $ DF \cong FG $.

### Step 2: Use Transitive Property of Congruence
- We have $ BC \cong AB $ (from the bisector definition) and $ FG \cong DF $ (from the bisector definition). Also, $ AB \cong FG $ (Given). By the transitive property of segment congruence (if $ a \cong b $ and $ b \cong c $, then $ a \cong c $), we can conclude $ BC \cong FG $ (since $ BC \cong AB $ and $ AB \cong FG $). Then, using the transitive property again with $ BC \cong FG $ and $ FG \cong DF $, we get $ BC \cong DF $.

### Filling in the Blanks:
1. **Step 2 (BF bisects...):** Since $ BF $ bisects $ AC $ and $ DG $, by the definition of a segment bisector, $ AB \cong BC $ and $ DF \cong FG $. So the blanks here are $ AC $ and $ DG $ (for what $ BF $ bisects), and later in step 3, we use the bisector definition to get $ BC \cong AB $ and $ FG \cong DF $.
2. **Step 3 (Definition of segment bisector):** The statements are $ BC \cong AB $ and $ FG \cong DF $ (from the bisector dividing $ AC $ into $ AB \cong BC $ and $ DG $ into $ DF \cong FG $).
3. **Step 4 (Transitive Property of Segment Congruence):** We use $ AB \cong FG $ (Given) and $ BC \cong AB $, $ FG \cong DF $ (from Step 3) to get $ BC \cong FG $ (transitive: $ BC \cong AB \cong FG $).
4. **Step 5 (Transitive Property or Direct Congruence):** Using $ BC \cong FG $ (Step 4) and $ FG \cong DF $ (Step 3), by transitive property, $ BC \cong DF $.

### Final Proof Structure (Filling Blanks):
- **Step 2:** $ BF $ bisects $ \boldsymbol{AC} $ and $ \boldsymbol{DG} $.
- **Step 3:** $ \boldsymbol{BC} \cong AB $, $ FG \cong \boldsymbol{DF} $ (Definition of segment bisector).
- **Step 4:** $ \boldsymbol{AB} \cong FG $, so $ BC \cong FG $ (Transitive Property of Segment Congruence).
- **Step 5:** $ BC \cong \boldsymbol{DF} $ (Transitive Property: $ BC \cong FG $ and $ FG \cong DF $, so $ BC \cong DF $).

### Key Justifications:
- **Definition of Segment Bisector:** A bisector splits a segment into two congruent parts.
- **Transitive Property of Segment Congruence:** If $ \overline{X} \cong \overline{Y} $ and $ \overline{Y} \cong \overline{Z} $, then $ \overline{X} \cong \overline{Z} $.

For the final answer (filling the last blank in Step 5):
$ BC \cong \boldsymbol{DF} $ (by transitive property, since $ BC \cong FG $ and $ FG \cong DF $).

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QUESTION IMAGE

Question

Explanation:

Step 1: Analyze Given Information

Step 2: Use Transitive Property of Congruence

Filling in the Blanks:

Final Proof Structure (Filling Blanks):

Key Justifications:

Answer:

Step 1: Analyze Given Information

Step 2: Use Transitive Property of Congruence

Filling in the Blanks:

Final Proof Structure (Filling Blanks):

Key Justifications: