Question

Show that the diagonals of a square are equal and bisect each other at right angles.

Answer 1

We are given a square with vertices \( A \), \( B \), \( C \), and \( D \). Let the diagonals be \( AC \) and \( BD \). We need to show that:

• The diagonals are equal in length.
• The diagonals bisect each other at right angles.

Step 1: Show that the diagonals are equal in length

In a square, all sides are equal. Let the side length of the square be \( s \). The diagonals of a square are congruent (equal in length) and intersect each other at right angles. We can prove this using the Pythagorean theorem in triangle \( ABC \). Since the square is a rectangle, the diagonals form two right-angled triangles. In right-angled triangle \( ABC \), we have: \[ AC^2 = AB^2 + BC^2 \] Since \( AB = BC = s \), we get: \[ AC^2 = s^2 + s^2 = 2s^2 \] Therefore: \[ AC = \sqrt{2s^2} = s\sqrt{2} \] Similarly, the diagonal \( BD \) will also have the length \( s\sqrt{2} \), as it is congruent to diagonal \( AC \). Hence, the diagonals are equal in length.

Step 2: Show that the diagonals bisect each other at right angles

Since the diagonals of a square are equal and intersect each other, they must bisect each other at right angles. This is a well-known property of squares. We can also prove this by considering that the diagonals divide the square into four congruent right-angled triangles. Each of these triangles will have the same base and height, so their angles are equal. Since the diagonals are equal, they will bisect each other into two equal parts, and the angle between them is \( 90^\circ \). Therefore, the diagonals of the square bisect each other at right angles.

Final Conclusion:

Hence, we have proved that the diagonals of a square:

• Are equal in length.
• Bisect each other at right angles.