To critique their reasoning and identify the best option to fill in the blank, we need to consider the principles of similar triangles. The statement that "corresponding sides of similar triangles are proportional" is central to proving that \( DF = \frac{1}{2} BC \).
Given that \( \triangle ADF \sim \triangle ABC \), we can write the ratio of corresponding sides:
\[ \frac{DF}{BC} = \frac{AD}{AB} = \frac{AF}{AC} \]
To conclude that \( DF = \frac{1}{2} BC \), we want to assert that:
\[ \frac{DF}{BC} = \frac{1}{2} \]
Thus, the missing portion of their proof should indicate that the ratio of the lengths corresponds to the relationship:
\[ \frac{DF}{BC} = \frac{1}{2} \]
This is captured by the option:
Start Fraction upper D upper F over upper B upper C End Fraction equals Start Fraction 1 over 2 End Fraction.
Therefore, the best option to fill in the blank is:
Start Fraction upper D upper F over upper B upper C End Fraction equals Start Fraction 1 over 2 End Fraction.
This confirms that \( DF = \frac{1}{2} BC \) and leads to the conclusion that \( DF \parallel BC \).