Invariance in Orthodiagonal Quadrilaterals
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A Mathematical Droodle

23 September 2015, Created with GeoGebra

Explanation

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Copyright © 1996-2017 Alexander Bogomolny

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.

23 September 2015, Created with GeoGebra

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,

which only makes sense if cosα = 0, i.e., when α = 90°. (We have this result geometrically as a part of Floor Lamoen's proof of the Pythagorean theorem.)

Note that this property is analogous to a similar property of inscriptible quadrilaterals that are characterized by the identity a + c = b + d. Displacements of the vertices of the latter that preserve the side lengths leave the quadrilateral inscriptible. Kites are both orthodiagonal and inscriptible.


Orthodiagonal Quadrilaterals

  1. Invariance in Orthodiagonal Quadrilaterals
  2. Orthodiagonal and Cyclic Quadrilaterals
  3. Classification of Quadrilaterals
  4. Pythagorean Theorem in an Orthodiagonal Quadrilateral
  5. Easy Construction of Bicentric Quadrilateral
  6. Easy Construction of Bicentric Quadrilateral II

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2017 Alexander Bogomolny

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