Question

Prove that diagonals of a rectangle are congruent.

Answer 1

To prove that the diagonals of a rectangle are congruent, we can use the properties of triangles and the definition of a rectangle.

### Definition and Properties
A rectangle is defined as a quadrilateral with four right angles. Let's denote the rectangle as \( ABCD \) with:
- \( A \) at the top-left,
- \( B \) at the top-right,
- \( C \) at the bottom-right, and
- \( D \) at the bottom-left.

### Diagonals
The diagonals of the rectangle are \( AC \) and \( BD \). We need to show that \( AC \cong BD \).

### Coordinate System
For simplicity, we can place the rectangle in a coordinate system:
- Let the coordinates of \( A \) be \( (0, 0) \),
- \( B \) be \( (a, 0) \),
- \( C \) be \( (a, b) \), and
- \( D \) be \( (0, b) \),
where \( a \) and \( b \) are the lengths of the sides of the rectangle.

### Calculate Lengths of Diagonals
1. **Length of Diagonal \( AC \)**:
\[
AC = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2} = \sqrt{(a - 0)^2 + (b - 0)^2} = \sqrt{a^2 + b^2}
\]

2. **Length of Diagonal \( BD \)**:
\[
BD = \sqrt{(x_D - x_B)^2 + (y_D - y_B)^2} = \sqrt{(0 - a)^2 + (b - 0)^2} = \sqrt{(-a)^2 + b^2} = \sqrt{a^2 + b^2}
\]

### Conclusion
Since both diagonals \( AC \) and \( BD \) have been calculated to have the same length:
\[
AC = BD = \sqrt{a^2 + b^2}
\]

Thus, we conclude that the diagonals of a rectangle are congruent:
\[
AC \cong BD
\]

This completes the proof.