Question

Given: ABCD is a rhombus and △ACB ≅ △DBC

Prove: ABCD is a square

Answer 1

1. Since ABCD is a rhombus, all sides are congruent.

2. Since △ACB ≅ △DBC, ∠ACB ≅ ∠DBC.

3. Since opposite angles of a parallelogram are congruent, ∠ABC ≅ ∠DCB.

4. Since ∠ACB ≅ ∠DBC and ∠ABC ≅ ∠DCB, then ∠ACB + ∠ABC = ∠DBC + ∠DCB.

5. Since the sum of the angles in a triangle is 180°, then ∠ACB + ∠ABC = 180° and ∠DBC + ∠DCB = 180°.

6. Therefore, ABCD is a rectangle.

7. Since ABCD is both a rhombus and a rectangle, it must be a square.

Answer 2

To prove that ABCD is a square, we need to show that all four angles of the rhombus are right angles. Given that △ACB ≅ △DBC, we can conclude that ABCD is a square.

Step-by-step explanation:

To prove that ABCD is a square, we need to show that all four angles of the rhombus are right angles.

Given that △ACB ≅ △DBC, we can conclude the following:

1. AC = DB (Corresponding parts of congruent triangles are congruent).
2. ∠ACB = ∠DBC (Corresponding parts of congruent triangles are congruent).

Since AC = DB and ∠ACB = ∠DBC, we can conclude that △ACB and △DBC are congruent.

Since opposite sides of a rhombus are parallel and congruent, and △ACB and △DBC are congruent, it follows that ABCD is a rectangle.

Further, since all four angles of a rhombus are congruent, if ABCD is a rectangle, then all four angles of ABCD are right angles. Therefore, ABCD is a square.