Any angle will do

Two things are settled: the diagonals must be equal and must bisect each other. What about the third question: does the angle between them matter?

Try one. Draw two equal diagonals bisecting each other at 60°. The four angles at O are then 60°, 60°, 120°, 120° (vertically opposite angles, and pairs on a straight line add to 180°).

In ∆AOB, OA = OB (each is half of an equal diagonal), so its two base angles are equal; call each a. Then a + a + 60 = 180, so 2a = 120 and a = 60. Work round the figure and every angle of the quadrilateral comes out as 30° + 60° = 90°.

One case proves nothing, so make it general. Let the angle be x. The four angles at O are x, x, 180 − x, 180 − x.

• In isosceles ∆AOB with base angles a: a + a + x = 180, so 2a = 180 − x and a = 90 − x/2.
• In isosceles ∆AOD with base angles b: b + b + (180 − x) = 180, so 2b = x and b = x/2.

Each angle of the quadrilateral is a + b = (90 − x/2) + x/2 = 90. The x cancels, which is exactly the point. And since ∆AOB ≅ ∆COD and ∆AOD ≅ ∆COB, AB = CD and AD = CB. So whatever the angle, the figure is a rectangle. (Make the angle 90° and you get a special one, the square; more on that soon.)

That's the carpenter's answer: equal strips, joined at their midpoints, at any angle at all. Carpenters in Europe use this to square a frame, and farmers in Mozambique use it to set out the rectangular base of a house.

And notice what kind of answer it is. Measuring a few rectangles would only give a conjecture: 'a statement about which we are highly confident, but not yet sure if it always holds true'. Can we be sure the 1000th rectangle behaves? Only a proof settles that.