Pythagorean Theorem in an Orthodiagonal Quadrilateral
Let there be two squares: APBMc and C1McC2Q with a common vertex Mc. Rotation through 90o in the clockwise direction around Mc, moves C2Mc into C1Mc and AMc into BMc. This implies that DAMcC2 rotates into DBMcC1 so that AC2 and BC1 are orthogonal. Quadrilateral ABC2C1 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 APBMc and C1McC2Q is half this quantity.
In a special case where Mc 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 APBMc and C1McC2Q add up to that of a red square!
|What if applet does not run?|
- 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