# Pythagorean Theorem in an Orthodiagonal Quadrilateral

Let there be two squares: APBM_{c} and C_{1}M_{c}C_{2}Q with a common vertex M_{c}. Rotation through 90^{o} in the clockwise direction around M_{c}, moves C_{2}M_{c} into C_{1}M_{c} and AM_{c} into BM_{c}. This implies that DAM_{c}C_{2} rotates into DBM_{c}C_{1} so that AC_{2} and BC_{1} are orthogonal. Quadrilateral ABC_{2}C_{1} is thus *orthodiagonal* (i.e., having orthogonal diagonals) and the configurations is remarkable: the red and blue squares add up to the same area. The important point to note is that the sum of the areas of the original squares APBM_{c} and C_{1}M_{c}C_{2}Q is half this quantity.

In a special case where M_{c} coincides with the point of intersection of the diagonals we get a proof of the Pythagorean theorem:

Because of the resulting symmetry, the red squares are equal. Therefore, the areas of APBM_{c} and C_{1}M_{c}C_{2}Q add up to that of a red square!

What if applet does not run? |

### Orthodiagonal Quadrilaterals

- Invariance in Orthodiagonal Quadrilaterals
- Orthodiagonal and Cyclic Quadrilaterals
- Classification of Quadrilaterals
- Pythagorean Theorem in an Orthodiagonal Quadrilateral
- Easy Construction of Bicentric Quadrilateral
- Easy Construction of Bicentric Quadrilateral II

|Up| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny69248240