Question

A line connects midpoints upper D and upper F. Segment upper A upper D and segment upper D upper B are marked with single congruent tick marks. Segment upper A upper F, and segment upper F upper C are marked with double congruent tick marks.


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

DF=BC

upper D upper F equals upper B upper C

BCDF=2

Answer 1

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

Comment: From the similarity DF/BC = AD/AB. Since AD = DB, AB = 2·AD so AD/AB = 1/2, hence DF/BC = 1/2 and multiplying by BC gives DF = (1/2)BC.