If ABCDE is a pentagon, then `bar(AB) + bar(AE) + bar(BC) + bar(DC) + bar(ED) =`

Options

A. `bar(AC)`
B. 2`bar(AC)`
C. 2`bar(AD)`
D. 3`bar(AD)`

Answer

To solve the problem, we need to analyze the vector equation given for the pentagon ABCDE. The expression we need to evaluate is:

\[ \overline{AB} + \overline{AE} + \overline{BC} + \overline{DC} + \overline{ED} \]

### Step-by-Step Solution:

1. **Identify the Vectors**:
- Let \(\overline{AB}\) be the vector from point A to point B.
- Let \(\overline{AE}\) be the vector from point A to point E.
- Let \(\overline{BC}\) be the vector from point B to point C.
- Let \(\overline{DC}\) be the vector from point D to point C.
- Let \(\overline{ED}\) be the vector from point E to point D.

2. **Rearranging the Vectors**:
- We can rearrange the expression to group the vectors that share common points:
\[ \overline{AB} + \overline{AE} + \overline{BC} + \overline{DC} + \overline{ED} = \overline{AB} + \overline{BC} + \overline{AE} + \overline{ED} + \overline{DC} \]

3. **Using Vector Addition**:
- Notice that \(\overline{AB} + \overline{BC} = \overline{AC}\) (the vector from A to C).
- Similarly, we can combine \(\overline{AE} + \overline{ED} = \overline{AD}\) (the vector from A to D).

4. **Combining the Results**:
- Now we can rewrite the expression:
\[ \overline{AC} + \overline{AD} + \overline{DC} \]

5. **Further Simplification**:
- The segment \(\overline{AD} + \overline{DC}\) can be combined as \(\overline{AC}\) since it connects points A to C through D.
- Therefore, we have:
\[ \overline{AC} + \overline{AC} = 2\overline{AC} \]

6. **Final Result**:
- Thus, the final result is:
\[ \overline{AB} + \overline{AE} + \overline{BC} + \overline{DC} + \overline{ED} = 2 \overline{AC} \]

### Conclusion:
The expression evaluates to \(2 \overline{AC}\).