Question

Given: A, B, C, DABCD is a rhombus and triangle, A, C, B, \cong, triangle, D, B, C, .△ACB≅△DBC.

Prove: A, B, C, DABCD is a square.
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
triangle, A, C, B, \cong, triangle, D, B, C△ACB≅△DBC
Given. What are the remaining steps for the question?

Answer 1

Fill in the following steps (statement — reason):

1. A, B, C, D; ABCD is a rhombus. Triangles ACB ≅ DBC. — Given.
2. AC = BD. — Corresponding parts of congruent triangles (CPCTC).
3. ABCD is a parallelogram. — Definition of rhombus (a rhombus is a parallelogram with all sides equal).
4. In any parallelogram, AC^2 + BD^2 = 2(AB^2 + BC^2). — Parallelogram law (relation between sides and diagonals).
5. Since AC = BD, 2 AC^2 = 2(AB^2 + BC^2), so AC^2 = AB^2 + BC^2. — Substitution / algebra.
6. Therefore ∠ABC = 90°. — Converse of the Pythagorean theorem (in triangle ABC).
7. ABCD is a rhombus with one right angle, so ABCD is a square. — A rhombus with a right angle is a rectangle with all sides equal, i.e. a square.

Thus ABCD is a square.