Opal has already proved that △ADF∼△ABC . They are now attempting to prove that DF=12BC and DF¯¯¯¯¯¯¯¯∥BC¯¯¯¯¯¯¯¯ . Critique their reasoning. Which option should fill in the blank?

Question

PROOF: Because corresponding sides of similar triangles are proportional, ________________. Multiply both sides of the equation by segment BC to show that DF=12BC . Because corresponding angles of similar triangles are congruent, ∠ADF≅∠ABC . If corresponding angles are congruent, then lines are parallel, so DF¯¯¯¯¯¯¯¯∥BC¯¯¯¯¯¯¯¯ .

(1 point)
Responses

DFBC=12
Start Fraction upper D upper F over upper B upper C End Fraction equals Start Fraction 1 over 2 End Fraction

DF=BC
upper D upper F equals upper B upper C

AD=DB
upper A upper D equals upper D upper B

BCDF=2
Start Fraction upper B upper C over upper D upper F End Fraction equals 2

Answer 1

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 \( DF \parallel BC \).