the diagonals of the square are perpendicular with each other. Prove analytically.?

Question

analytic proof

Answer 1

I don't know what is analatycal proof, but I'll solve it my own way. I hope it helps.
A square is a rhombus with a right angle. The diagonals are equal and bisect each other.
To prove:The diagonals of the square are perpendicular to each other.
Take a square ABCD
Diagonals AC and BD intersect at O.
AB = BC = CD = AD
AC = BD
AO = OC = BO = OD
So, all the triangles formed by the diagonals intersecting are congruent.
So,
Angle AOB = Angle BOC = Angle COD = Angle DOA
But,
Angle AOB + Angle BOC + Angle COD + Angle DOA = 360 degrees
4(Angle AOB) = 360
Angle AOB = 90
Hence proved

Answer 2

Using Analytic Geometry, you can choose your axes to suit the figure. So, start off with any square, choose the bottom left corner as origin, and the two sides through it as axes. You can also choose the unit, so make the length of the side of the square the unit.
Then the coordinates of the vertices are O(0,0), A(1,0), B(1,1) and C(0,1).
Use the gradient formula to get the gradients of OB and CA, multiply them together and find that you get -1, which proves that they are perpendicular to each other.

Answer 3

I'll prove it with vectors: The diagonals are AC=(t,t) and BD=(t,-t) The dot product of AC and BD is t(t)+(t)(-t)=0. Since AC is orthogonal to BD if and only if the dot product equals 0, AC is orthogonal to BD, which means AC is perpendicular to BD. Thus, the diagonals of a square are perpendicular.