Question

Fill in the missing statement and reason of the proof below.

Given: A, B, C, DABCD is a parallelogram and DD is the midpoint of start overline, A, E, end overline, .
AE
.

Prove: start overline, B, D, end overline, \cong, start overline, C, E, end overline
BD
≅
CE
.
In the proof shown in the table below, one of the steps in the middle is missing. Before completing the missing step, make sure to read the steps that come after it. To complete the missing step, first use the dropdown menu in the statement column to select the format of the statement for that step. Once you select a format, a statement will appear with input boxes and/or dropdowns to complete, and a dropdown menu will appear in the reason column. Once completed, press the submit button in the answer area at the bottom of the page. The diagram associated with this problem is located below the proof table, 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
DD is the midpoint of start overline, A, E, end overline
AE

Given
2
start overline, A, D, end overline, \cong, start overline, D, E, end overline
AD
≅
DE

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

Opposite sides of a parallelogram are congruent
4
start overline, D, E, end overline, \cong, start overline, B, C, end overline
DE
≅
BC

Transitive Property
5
start overline, C, D, end overline, \cong, start overline, C, D, end overline
CD
≅
CD

Reflexive Property
6
start overline, B, C, end overline, \parallel, start overline, A, D, end overline
BC
∥
AD

Opposite sides of a parallelogram are parallel
7 type a statement
8
triangle, B, C, D, \cong, triangle, E, D, C△BCD≅△EDC
SAS
9
start overline, B, D, end overline, \cong, start overline, C, E, end overline
BD
≅
CE

Corresponding Parts of Congruent Triangles are Congruent (CPCTC)

Answer 1

Step 7 statement: ∠BCD ≅ ∠EDC

Reason: BC ∥ AD and A, D, E are collinear (D is midpoint of AE), so BC ∥ DE; parallel lines cut by transversal CD give corresponding (or alternate interior) angles congruent.