Invariance in Orthodiagonal Quadrilaterals
What Is This About?
A Mathematical Droodle

The applet attempts to suggest a certain property of orthodiagonal quadrilaterals. (A non-degenerate quadrilateral ABCD is orthodiagonal if AC ⊥ BD.)

Let ABCD be an orthodiagonal quadrilateral; let A'B'C'D' be a quadrilateral with the same sides: A'B' = AB, B'C' = BC, C'D' = CD, D'A' = DA. Then A'B'C'D' is also orthodiagonal.

Put differently, imagine that material segments AB, BC, CD, DA are hinged at the vertices A, B, C, D. The statement asserts that, under any displacement of the vertices that preserves the side lengths, an orthodiagonal quadrilateral will remain orthodiagonal.

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Proof

Let, for simplicity, AB = a, BC = b, CD = c, DA = d. We'll show that

A quadrilateral ABCD is orthodiagonal iff a² + c² = b² + d².

Let P be the point of intersection of the diagonals AC and BD. Denote x = AP, y = BP, z = CP, w = DP.

By the Pythagorean theorem, in an orthodiagonal quadrilateral ABCD,

x² + y² = a²,
y² + z² = b²,
z² + w² = c²,
w² + x² = d².

Adding the first identity to the third and the second to the fourth, we observe that the left-hand side is in both cases x² + y² + z² + w². Thus the right-hand sides are also equal: a² + c² = b² + d².

For an arbitrary quadrilateral ABCD, let α = ∠APB. In this case we apply the Law of Cosines:

a² = x² + y² - 2xy cosα,
b² = y² + z² + 2yz cosα,
c² = z² + w² - 2zw cosα,
d² = w² + x² + 2wx cosα.

Again adding the first identity to the third and the second to the fourth, we get, after cancelling out the squares:

(x + z)(y + w) cosα = 0,