Write a formal proof of the theorem.?

Question

The diagonals of a square are perpendicular.

Answer 1

Angus gave a proof out of analytical geometry. Not bad.

Sometimes it isn't clear what the ground rules are; I'll give you another proof that doesn't use coordinates (although Angus' will be easier to read, I think).

Call the vertices A, B, C, and D as you go around the square; let E be the point of intersection of the diagonals. Since ABCD is a square, AB=BC=CD=AD and angles ABC, BCD, CDA, and DAB are all right angles. This means that triangles ACD and BCD are isosceles right triangles (45-45-90), so angles BDC and ACD are each 45 degrees. Those two angles are also known by the names EDC and ECD. Since the interior angles of a triangle add up to 180 degrees, this means that angle CED is a right angle, so the diagonals are perpendicular.

Answer 2

This is a difficult question because it is not clear what you have at your disposal. One simple method is simply from the definition of a square as a rhombus and a rectangle. And, the diagonals of a rhombus are perpendicular.

If your definition of a square is all sides congruent and all angles right angles, then you can build up the properties of the diagonals. If you have already proven that the diagonals of a parallelogram bisect each other, and those of a rectangle are congruent, then the diagonals of the square have these properties. So all four of the triangles created by the diagonals are congruent by SSS. Then their angles are right angles, as they are congruent and supplementary.

But that depends on what you have already proven!

Answer 3

Let A, B, C and D be the vertices of the square.
locate the points on a cartesian plane, such that A(0,a), B(a,a), C
(a,0) and D(0,0).
the diagonals are then lines AC and BD.

find the slope of AC = (0-a)/(a-0)= -1
find the slope of BD = (a-0)/(a-0)= 1

-1 is the negative reciprocal of 1 hence the lines AC and BD are perpendicular.