Answer:
To prove that the diagonals of a square are equal and bisect each other at a right angle, we can use the following steps:Let ABCD be a square, with diagonal AC and BD.First, we can show that the diagonals are equal:Using the Pythagorean theorem, we can find the length of AC and BD in terms of the side length of the square, which we can denote as s:AC^2 = AB^2 + BC^2 (by the Pythagorean theorem)
AC^2 = s^2 + s^2 (since AB = BC = s in a square)
AC^2 = 2s^2
AC = sqrt(2s^2)Similarly, we can find the length of BD:BD^2 = BA^2 + AD^2 (by the Pythagorean theorem)
BD^2 = s^2 + s^2 (since BA = AD = s in a square)
BD^2 = 2s^2
BD = sqrt(2s^2)Since AC and BD are both equal to sqrt(2s^2), we have shown that the diagonals of a square are equal.Next, we can show that the diagonals bisect each other at a right angle:Let M be the midpoint of AC, and let N be the midpoint of BD. Since M is the midpoint of AC, we have:AM = MC = sqrt(2s^2)/2Similarly, we can find that:BN = ND = sqrt(2s^2)/2Since AM = MC and BN = ND, we know that M and N lie on the perpendicular bisectors of AC and BD, respectively.To show that the diagonals bisect each other at a right angle, we just need to show that the line segments MN and BD are perpendicular. We can do this by showing that the slopes of these lines are negative reciprocals of each other:The slope of BD is (0 - s)/(s - 0) = -1.The slope of MN can be found by noting that the line segment MN is parallel to AB and CD (since M and N are midpoints). The slope of AB is (s - 0)/(0 - s) = -1, so the slope of MN is also -1.Since the slopes of MN and BD are negative reciprocals of each other, we know that these lines are perpendicular.Therefore, we have shown that the diagonals of a square are equal and bisect each other at a right angle.