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

To demonstrate that the diagonals of a square are equal and bisect each other at right angles, let's consider a square ABCD with vertices A, B, C, and D.

Equal Length of Diagonals

The diagonals of the square are AC and BD. Since all sides of a square are equal, we can use the Pythagorean theorem to find the lengths of the diagonals.

• In triangle ABC, the sides AB and BC are equal to the side length of the square, which we can denote as 's'.
• The diagonal AC can be calculated as follows:

AC = √(AB² + BC²) = √(s² + s²) = √(2s²) = s√2.

Similarly, for diagonal BD, using the same reasoning:

BD = √(AD² + DC²) = √(s² + s²) = s√2.

Thus, we find that AC = BD, proving that the diagonals are equal.

Bisecting Each Other at Right Angles

Next, we need to show that the diagonals bisect each other at right angles. Let’s denote the intersection point of the diagonals as O.

• Since the diagonals of a square are symmetrical, point O is the midpoint of both AC and BD.
• Each diagonal divides the square into two congruent triangles.

To show that they intersect at right angles, consider triangles AOB and COD:

• In triangle AOB, angle AOB is formed by the diagonals.
• Since triangle AOB and triangle COD are congruent, angle AOB is equal to angle COD.

Since the sum of angles in a triangle is 180 degrees, and both triangles are right triangles, we conclude that:

Angle AOB + Angle BOC + Angle COD = 180 degrees.

Given that angle AOB and angle COD are equal, it follows that:

Angle AOB = Angle COD = 90 degrees.

Final Summary

In summary, the diagonals of a square are equal in length and intersect at right angles, bisecting each other at point O. This property is a fundamental characteristic of squares, showcasing their symmetry and geometric elegance.