Asked by Sam

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

Given: start overline, A, C, end overline
AC
and start overline, B, D, end overline
BD
bisect each other and start overline, B, F, end overline, \cong, start overline, D, G, end overline, .
BF
≅
DG
.

Prove: start overline, F, E, end overline, \cong, start overline, G, E, end overline
FE
≅
GE
.
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
start overline, A, C, end overline
AC
and start overline, B, D, end overline
BD
bisect each other
start overline, B, F, end overline, \cong, start overline, D, G, end overline
BF
≅
DG

Given
2
A, B, C, DABCD is a parallelogram
A, B, C, DABCD is a parallelogram because it is a quadrilateral with diagonals that bisect each other
3
start overline, A, B, end overline, \parallel, start overline, C, D, end overline
AB
∥
CD

Opposite sides of a parallelogram are parallel
4 type of statement
5
start overline, B, E, end overline, \cong, start overline, D, E, end overline
BE
≅
DE

A segment bisector divides a segment into two congruent segments
6
triangle, F, B, E, \cong, triangle, G, D, E△FBE≅△GDE
SAS
7
start overline, F, E, end overline, \cong, start overline, G, E, end overline
FE
≅
GE

Corresponding Parts of Congruent Triangles are Congruent (CPCTC)

All Answers 1

Answered by GPT-5 mini AI
Step 4 statement: ∠FBE ≅ ∠GDE

Reason: If two parallel lines are cut by a transversal, corresponding (or alternate interior) angles are congruent (here AB ∥ CD and BE is the transversal).