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Given: angle, E, C, F, \cong, angle, C, F, D, comma∠ECF≅∠CFD, start overline, A, B, end overline, \parallel, start overline, F, C, end overline, comma
AB
∥
FC
, start overline, E, B, end overline, \cong, start overline, F, D, end overline
EB
≅
FD
and angle, B, \cong, angle, D, .∠B≅∠D.
Prove: A, E, C, FAECF is a rhombus.
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
angle, E, C, F, \cong, angle, C, F, D∠ECF≅∠CFD
start overline, A, B, end overline, \parallel, start overline, F, C, end overline
AB
∥
FC
start overline, E, B, end overline, \cong, start overline, F, D, end overline
EB
≅
FD
angle, B, \cong, angle, D∠B≅∠D
Given
2
start overline, E, C, end overline, \parallel, start overline, A, D, end overline
EC
∥
AD
If two lines cut by a transversal form congruent alternate interior angles, then the two lines are parallel
3
A, E, C, FAECF is a parallelogram
A, E, C, FAECF is a parallelogram because it is a
quadrilateral
with
one pair of opposite sides parallel
4
angle, B, E, C, \cong, angle, E, C, F∠BEC≅∠ECF
Corresponding Parts of Congruent Triangles are Congruent (CPCTC)
6
triangle, E, B, C, \cong, triangle, F, D, C△EBC≅△FDC
SAS
7
start overline, E, C, end overline, \cong, start overline, F, C, end overline
EC
≅
FC
In a triangle, sides opposite of congruent angles are congruent
8
A, E, C, FAECF is a rhombus
A, E, C, FAECF is a rhombus because it is a
quadrilateral
with
one pair of opposite sides parallel
The proof is complete. You may now press submit. You can still change reasons.
answer are reflexive property transitive porperty substitution corresponding parts of congruent triangles are congruent (CPCTC) in a triangle, sides are opposite of the congruent angles are congruent a segment bisector divides a segment into two congruent segments a midpoint divides a segment into two congruent segments opposit e sides of a parallelogram are congruent the diagonal of a rectangle/square are congruent all sides of a rhombus/square are congruent congruent segments added to congruent segments form congreunt segments congreunt segments subtracted from congruent segments form congruent segments segments that are half the lengths of congruent segments are congruent segments that are twice the lengths of congruent segments are congruent sss sas asa aas ssa aa h
3 answers
To complete the proof that quadrilateral \( AECF \) is a rhombus using the provided steps, let's fill in the necessary reasons for each statement as outlined in your proof table:
Given: \( \angle ECF \cong \angle CFD \), \( AB \parallel FC \), \( EB \cong FD \), \( \angle B \cong \angle D \)
Reason: Given
\( EC \parallel AD \)
Reason: If two lines cut by a transversal form congruent alternate interior angles, then the two lines are parallel.
\( AECF \) is a parallelogram.
Reason: AECF is a parallelogram because it is a quadrilateral with one pair of opposite sides parallel.
\( \angle BEC \cong \angle ECF \)
Reason: Reflexive Property
\( \angle BEC \cong \angle DFC \)
Reason: Corresponding Parts of Congruent Triangles are Congruent (CPCTC)
\( \triangle EBC \cong \triangle FDC \)
Reason: SAS
\( EC \cong FC \)
Reason: In a triangle, sides opposite of congruent angles are congruent.
\( AECF \) is a rhombus.
Reason: AECF is a rhombus because it is a quadrilateral with all sides congruent (which implies opposite sides are equal).
With all the steps clarified, the proof successfully concludes that \( AECF \) is a rhombus based on the information and reasoning provided.
After verifying all statements and reasons, you can now submit the completed proof.
same answers Given: A, B, C, DABCD is a parallelogram and start overline, B, E, end overline, \parallel, start overline, F, D, end overline, .
BE
∥
FD
.
Prove: triangle, A, B, E, \cong, triangle, C, D, F△ABE≅△CDF.
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
start overline, B, E, end overline, \parallel, start overline, F, D, end overline
BE
∥
FD
Given
2
angle, B, E, A, \cong, angle, D, F, C∠BEA≅∠DFC
Reflexive Property
3
start overline, A, B, end overline, \cong, start overline, C, D, end overline
AB
≅
CD
Transitive Property
4
start overline, A, B, end overline, \parallel, start overline, C, D, end overline
AB
∥
CD
If two lines cut by a transversal form congruent alternate interior angles, then the two lines are parallel
5
angle, B, A, E, \cong, angle, D, C, F∠BAE≅∠DCF
Vertical angles are congruent
6
triangle, A, B, E, \cong, triangle, C, D, F△ABE≅△CDF
AAS
The proof is complete. You may now press submit. You can still change reasons.
A
B
C
D
E
F
An image is shown. To download tactile, press SHIFT-CONTROL-U for UEB or SHIFT-CONTROL-E for Nemeth.
Note: the segment A, CAC is a straight segment.
Answer
Attempt 1 out of 2
To complete the proof that triangles \( ABE \) and \( CDF \) are congruent (\( \triangle ABE \cong \triangle CDF \)), we need to fill in the necessary reasons for each of the steps you provided. Based on your outline, here is the structured proof along with the selected reasons:
Given: \( ABCD \) is a parallelogram and \( BE \parallel FD \)
Reason: Given
\( \angle BEA \cong \angle DFC \)
Reason: Alternate Interior Angles are Congruent (since \( BE \parallel FD \) and \( AD \) serves as a transversal).
\( AB \cong CD \)
Reason: Opposite sides of a parallelogram are congruent.
\( AB \parallel CD \)
Reason: Opposite sides of a parallelogram are parallel.
\( \angle BAE \cong \angle DCF \)
Reason: Vertical angles are congruent (formed by the intersection of lines).
\( \triangle ABE \cong \triangle CDF \)
Reason: AAS (Angle-Angle-Side) - We have two angles and the non-included side congruent.
With these steps completed and the appropriate reasons documented, the proof establishes that \( \triangle ABE \cong \triangle CDF \).
You can go ahead and submit this completed proof after ensuring all statements and reasons are correctly filled in. Good luck!
1 answer