Let's analyze Monty's reasoning step by step and evaluate which option appropriately completes his argument regarding the similar triangles \( \triangle ADF \sim \triangle ABC \).
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Understanding Similar Triangles: Due to the similarity \(\triangle ADF \sim \triangle ABC\), we know:
- Corresponding sides are proportional.
- Corresponding angles are congruent.
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Proportionality Statement: From the similarity statement, it follows that: \[ \frac{DF}{BC} = \frac{AD}{AB} = \frac{AF}{AC} \] Therefore, Monty can conclude that: \[ DF = k \cdot BC \quad \text{(for some proportionality constant \( k \))} \]
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Claim of \( DF = \frac{1}{2} BC \): For Monty to state \( DF = \frac{1}{2} BC \), he must execute proper reasoning to support this specific proportionality. His equation \( DF BC = \frac{1}{2} \) (as written) contains a mistake in formatting — it should rather represent the ratio of sides.
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Argument about Parallel Lines: To conclude that \( DF \parallel BC \), Monty must establish the connection between congruent angles and parallel lines. This is captured by corresponding angles being equal when two lines are cut by a transversal (implying parallelism).
Now, let's analyze the provided options:
- Option #1: Because corresponding angles of similar triangles are congruent, \( \angle ADF \cong \angle ABC \) - This is correct regarding similarity but does not discuss parallel lines.
- Option #2: Because corresponding angles of similar triangles are congruent, \( \angle ADF \cong \angle DAF \) - This is incorrect since \( \angle DAF \) is not corresponding to \( \angle ADF \) in triangle \( ABC \).
- Option #3: Because corresponding angles are congruent when lines are parallel, \( \angle ADF \cong \angle ABC \) - This relates to angles and indicates parallel lines; however, it incorrectly states the source of congruence.
- Option #4: Because corresponding angles are congruent when lines are parallel, \( \angle ADF \cong \angle DAF \) - Incorrect, as the angles referenced are not corresponding angles in the context.
Best Choice: Option #1 is the most accurate and relevant option to go into the blank, even though it could use clearer phrasing about establishing \( DF \parallel BC \) based on angle equivalency from the similar triangles:
Final Critique
Final justification should clearly state that if \( \angle ADF \cong \angle ABC \) (from triangle similarity), we can also conclude that \( DF \parallel BC \) based on the properties of corresponding angles being equal indicating parallel lines crossed by a transversal.
So the completed proof should read: "Because corresponding angles of similar triangles are congruent, \( \angle ADF \cong \angle ABC \). If corresponding angles are congruent, then lines are parallel, so \( DF \parallel BC \)."
Thus, Monty’s argument, bolstered by Option #1, would be more robust.