Question

Use the image to answer the question.

A triangle is marked clockwise from the lower left vertex as upper A upper B upper C. Point upper D is marked at the midpoint of side upper A upper B. Point upper F is marked at the midpoint of side upper A upper C.

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?

PROOF: Because corresponding sides of similar triangles are proportional, ________________Blank. 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

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

DF=BC

Answer 1

Answer: Start Fraction DF over BC End Fraction = Start Fraction 1 over 2 End Fraction.

Reason: From △ADF ∼ △ABC we get DF/BC = AD/AB. Since D is the midpoint, AD/AB = 1/2, so DF/BC = 1/2. Multiplying by BC gives DF = (1/2)BC. The rest (angle congruence → parallel) is valid.