Question

Given: A, B, C, DABCD is a rhombus.

Prove: triangle, A, E, B△AEB is a right triangle.
In the table below, the first step of the proof is completed for you. To complete each following step, first use the dropdown menu in the "Statement" column to select the format of a statement. 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 you complete these inputs and dropdowns, navigate back to the first column and press the "try" button. If your statement and reason are valid, a new row will appear below. This process will continue until you have filled in a statement and reason for what you were asked to prove, and then the "try" button will cause the answer area to appear with a submit button. 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 rhombus
Given
2
start overline, A, B, end overline, \cong, start overline, B, C, end overline
AB
≅
BC

All sides of a rhombus are congruent
3
start overline, A, D, end overline, \cong, start overline, D, C, end overline
AD
≅
DC

All sides of a rhombus are congruent
4
start overline, A, B, end overline, \cong, start overline, A, D, end overline
AB
≅
AD

All sides of a rhombus are congruent
5
start overline, A, B, end overline, \parallel, start overline, C, D, end overline
AB
∥
CD

Opposite sides of a rhombus are parallel
6
start overline, B, C, end overline, \parallel, start overline, A, D, end overline
BC
∥
AD

Opposite sides of a rhombus are parallel
7
start overline, A, D, end overline, \cong, start overline, B, C, end overline
AD
≅
BC

Transitive Property
8
start overline, C, D, end overline, \cong, start overline, A, B, end overline
CD
≅
AB

Transitive Property. What are the last remaining steps, statements, and reasons for this question and please be more specific.

Answer 1

Continue from step 8 as follows (I number the new steps starting at 9 to match your table):

9. Statement: A, E, C are collinear and B, E, D are collinear.
Reason: E is the intersection of diagonals AC and BD (definition / from the diagram).

10. Statement: AE ≅ EC.
Reason: Diagonals of a parallelogram bisect each other. (Since AB ∥ CD and BC ∥ AD from steps 5 and 6, ABCD is a parallelogram, so its diagonals AC and BD bisect each other.)

11. (optional) Statement: BE ≅ ED.
Reason: Diagonals of a parallelogram bisect each other. (Same justification as step 10.)

12. Statement: ΔAEB ≅ ΔCEB.
Reason: SSS. AE ≅ CE (step 10), AB ≅ CB (step 2), and BE ≅ BE (common side).

13. Statement: ∠AEB ≅ ∠CEB.
Reason: CPCTC (corresponding parts of congruent triangles, from step 12).

14. Statement: ∠AEB + ∠CEB = 180°.
Reason: A, E, C are collinear (step 9) so ∠AEB and ∠CEB form a linear pair (supplementary).

15. Statement: ∠AEB = 90°.
Reason: If two equal angles are supplementary, each is a right angle (from steps 13 and 14: 2·∠AEB = 180°, so ∠AEB = 90°).

16. Statement: ΔAEB is a right triangle.
Reason: A triangle with a 90° angle is a right triangle (definition).

These steps show ∠AEB = 90°, so triangle AEB is a right triangle.