Question

Given: A, B, C, DABCD is a parallelogram and BB is the midpoint of start overline, A, F, end overline, .
AF
.

Prove: start overline, D, F, end overline
DF
bisects start overline, B, C, end overline
BC
.
In the proof shown in the table below, the statements and the first reason are filled in for you. Use the dropdown menu in the Reason column to select the reason for the next statement. Once you have selected a reason, a new statement will appear in a new row below. This will continue until you have selected the reason for the final statement, and then the answer area will appear with a submit button. The diagram associated with this problem is located at the bottom of the page, below the proof, and is available as a tactile printout. There may be a clarifying note below the diagram.
Step Statement Reason
1
A, B, C, DABCD is a parallelogram
BB is the midpoint of start overline, A, F, end overline
AF

Given
2
start overline, A, B, end overline, \cong, start overline, B, F, end overline
AB
≅
BF


A midpoint divides a segment into two congruent segments
3
start overline, A, B, end overline, \cong, start overline, D, C, end overline
AB
≅
DC


Opposite sides of a parallelogram / rhombus / rectangle / square are congruent
4
start overline, B, F, end overline, \cong, start overline, D, C, end overline
BF
≅
DC


In a triangle, sides opposite of congruent angles are congruent
5
start overline, D, C, end overline, \parallel, start overline, A, B, end overline
DC
∥
AB

Answer 1

Choose these reasons in order to complete the proof:

Step 4 (BF ≅ DC): Transitive property of congruence (if two segments are congruent to the same segment, they are congruent to each other).

Step 5 (DC ∥ AB): Opposite sides of a parallelogram are parallel.

Then continue with:

Step 6 (A, B, F are collinear): Definition of midpoint (B is the midpoint of AF).

Step 7 (BF ∥ DC): If A, B, F are collinear and AB ∥ DC, then BF ∥ DC.

Step 8 (BFCD is a parallelogram): If one pair of opposite sides of a quadrilateral are both congruent and parallel, the quadrilateral is a parallelogram.

Step 9 (DF bisects BC): Diagonals of a parallelogram bisect each other.

Thus DF bisects BC.