Question
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
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: start overline, A, B, end overline
AB
and start overline, A, D, end overline
AD
are segments.
Answer
Attempt 1 out of 2
You must answer all questions above in order to submit.
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
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: start overline, A, B, end overline
AB
and start overline, A, D, end overline
AD
are segments.
Answer
Attempt 1 out of 2
You must answer all questions above in order to submit.
Answers
Answered by
GPT-4o mini
To complete the proof, you will need to fill in the Statements and Reasons based on the information provided.
Let's continue the proof from Step 2 onwards.
Here are the steps and corresponding reasons filled out:
Step 1:
Statement: ∠ECF ≅ ∠CFD, AB ∥ FC, EB ≅ FD, ∠B ≅ ∠D
Reason: Given
Step 2:
Statement: EC ∥ AD
Reason: Corresponding Angles Postulate (since AB ∥ FC and angles ECF and CFD are corresponding angles)
Step 3:
Statement: EB || AD
Reason: Converse of Corresponding Angles Postulate (as angles ECF ≅ CFD and the lines are parallel)
Step 4:
Statement: AE ≅ CF
Reason: Congruent Triangles (by AAS or ASA, depending on which triangles you are referencing with the angles and sides you've proven to be congruent)
Step 5:
Statement: AECF is a rhombus
Reason: A rhombus has all sides equal in length.
Now let's compile the completed table:
| Step | Statement | Reason |
|------|------------------------------------------------|---------------------------------------------------|
| 1 | ∠ECF ≅ ∠CFD, AB ∥ FC, EB ≅ FD, ∠B ≅ ∠D | Given |
| 2 | EC ∥ AD | Corresponding Angles Postulate |
| 3 | EB || AD | Converse of Corresponding Angles Postulate |
| 4 | AE ≅ CF | Congruent Triangles (AAS or ASA) |
| 5 | AECF is a rhombus | A rhombus has all sides equal in length. |
Now you should be ready to submit your proof that AECF is a rhombus!
Let's continue the proof from Step 2 onwards.
Here are the steps and corresponding reasons filled out:
Step 1:
Statement: ∠ECF ≅ ∠CFD, AB ∥ FC, EB ≅ FD, ∠B ≅ ∠D
Reason: Given
Step 2:
Statement: EC ∥ AD
Reason: Corresponding Angles Postulate (since AB ∥ FC and angles ECF and CFD are corresponding angles)
Step 3:
Statement: EB || AD
Reason: Converse of Corresponding Angles Postulate (as angles ECF ≅ CFD and the lines are parallel)
Step 4:
Statement: AE ≅ CF
Reason: Congruent Triangles (by AAS or ASA, depending on which triangles you are referencing with the angles and sides you've proven to be congruent)
Step 5:
Statement: AECF is a rhombus
Reason: A rhombus has all sides equal in length.
Now let's compile the completed table:
| Step | Statement | Reason |
|------|------------------------------------------------|---------------------------------------------------|
| 1 | ∠ECF ≅ ∠CFD, AB ∥ FC, EB ≅ FD, ∠B ≅ ∠D | Given |
| 2 | EC ∥ AD | Corresponding Angles Postulate |
| 3 | EB || AD | Converse of Corresponding Angles Postulate |
| 4 | AE ≅ CF | Congruent Triangles (AAS or ASA) |
| 5 | AECF is a rhombus | A rhombus has all sides equal in length. |
Now you should be ready to submit your proof that AECF is a rhombus!
Answer
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 corresponding 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
four congruent sides
4
angle, B, E, C, \cong, angle, E, C, F∠BEC≅∠ECF
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: start overline, A, B, end overline
AB
and start overline, A, D, end overline
AD
are segments.
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 corresponding 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
four congruent sides
4
angle, B, E, C, \cong, angle, E, C, F∠BEC≅∠ECF
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: start overline, A, B, end overline
AB
and start overline, A, D, end overline
AD
are segments.
Answered by
GPT-4o mini
To complete the proof that AECF is a rhombus, we will fill in the statements and reasons in the provided table format.
Here is the completed proof with the statements you have and the filled-in reasons:
| Step | Statement | Reason |
|------|-------------------------------------------------|--------------------------------------------------------------------------------------------|
| 1 | ∠ECF ≅ ∠CFD, AB ∥ FC, EB ≅ FD, ∠B ≅ ∠D | Given |
| 2 | EC ∥ AD | If two lines cut by a transversal form congruent corresponding angles, then the two lines are parallel |
| 3 | AECF is a parallelogram | AECF is a parallelogram because it is a quadrilateral with four congruent sides |
| 4 | ∠BEC ≅ ∠ECF | Alternate Interior Angles Theorem (since EC || AD and EB || FD) |
| 5 | AECF is a rhombus | A rhombus is defined as a parallelogram with four congruent sides |
If you are asked to prove that AECF is a rhombus based on the statements provided above, the steps taken show that all sides are congruent and that opposite sides are parallel, satisfying the properties of a rhombus.
Now you can submit your proof!
Here is the completed proof with the statements you have and the filled-in reasons:
| Step | Statement | Reason |
|------|-------------------------------------------------|--------------------------------------------------------------------------------------------|
| 1 | ∠ECF ≅ ∠CFD, AB ∥ FC, EB ≅ FD, ∠B ≅ ∠D | Given |
| 2 | EC ∥ AD | If two lines cut by a transversal form congruent corresponding angles, then the two lines are parallel |
| 3 | AECF is a parallelogram | AECF is a parallelogram because it is a quadrilateral with four congruent sides |
| 4 | ∠BEC ≅ ∠ECF | Alternate Interior Angles Theorem (since EC || AD and EB || FD) |
| 5 | AECF is a rhombus | A rhombus is defined as a parallelogram with four congruent sides |
If you are asked to prove that AECF is a rhombus based on the statements provided above, the steps taken show that all sides are congruent and that opposite sides are parallel, satisfying the properties of a rhombus.
Now you can submit your proof!
Answer
what about angle bec = angle dfc
Answered by
GPT-4o mini
Certainly! Including the equality of angles BEC and DFC is an important detail. Here's how you can modify the proof steps to include that:
| Step | Statement | Reason |
|------|-------------------------------------------------|-----------------------------------------------------------------------------------|
| 1 | ∠ECF ≅ ∠CFD, AB ∥ FC, EB ≅ FD, ∠B ≅ ∠D | Given |
| 2 | EC ∥ AD | If two lines cut by a transversal form congruent corresponding angles, then the two lines are parallel |
| 3 | AECF is a parallelogram | AECF is a parallelogram because it is a quadrilateral with four congruent sides |
| 4 | ∠BEC ≅ ∠ECF and ∠DFC ≅ ∠ECF | Alternate Interior Angles Theorem (since EC || AD and AB || FC) |
| 5 | ∠BEC ≅ ∠DFC | Vertical Angles Theorem (∠BEC and ∠DFC are vertical angles) |
| 6 | AECF is a rhombus | A rhombus is defined as a parallelogram with four congruent sides (AE = EC = CF = FB) |
### Explanation of Steps
- **Step 4** shows that angles BEC and DFC are congruent because they are alternate interior angles formed by the transversal (EB and AD) intersecting the parallel lines.
- **Step 5** establishes that angles BEC and DFC are also equal because they are vertical angles.
Here you have a complete proof stating that AECF is a rhombus by showing it has four congruent sides and that the angles at the vertices fulfill the criteria of a rhombus.
Feel free to adjust based on your layout or additional criteria that the problem might specify!
| Step | Statement | Reason |
|------|-------------------------------------------------|-----------------------------------------------------------------------------------|
| 1 | ∠ECF ≅ ∠CFD, AB ∥ FC, EB ≅ FD, ∠B ≅ ∠D | Given |
| 2 | EC ∥ AD | If two lines cut by a transversal form congruent corresponding angles, then the two lines are parallel |
| 3 | AECF is a parallelogram | AECF is a parallelogram because it is a quadrilateral with four congruent sides |
| 4 | ∠BEC ≅ ∠ECF and ∠DFC ≅ ∠ECF | Alternate Interior Angles Theorem (since EC || AD and AB || FC) |
| 5 | ∠BEC ≅ ∠DFC | Vertical Angles Theorem (∠BEC and ∠DFC are vertical angles) |
| 6 | AECF is a rhombus | A rhombus is defined as a parallelogram with four congruent sides (AE = EC = CF = FB) |
### Explanation of Steps
- **Step 4** shows that angles BEC and DFC are congruent because they are alternate interior angles formed by the transversal (EB and AD) intersecting the parallel lines.
- **Step 5** establishes that angles BEC and DFC are also equal because they are vertical angles.
Here you have a complete proof stating that AECF is a rhombus by showing it has four congruent sides and that the angles at the vertices fulfill the criteria of a rhombus.
Feel free to adjust based on your layout or additional criteria that the problem might specify!