# Orthodiagonal quadrangles and Pythagoras

We prove in a simple way that the two sums of squares sidelengths of opposite
sides in an orthodiagonal quadrangle are equal. This implies the Pythagorean
theorem in two ways.Nederlandse
versie van deze pagina

#### Friendship of orthocenter and centroid as a lemma.

Consider two nonoverlapping squares attached to each other at a vertex. From
the common vertex two triangles are enclosed by these squares.

It is easy to see that an altitude from the common vertex in one triangle
extends to a median in the other triangle.

The median M_{c}R in the figure at the left hand side becomes a midparallel in
the figure at the right hand side, by rotation of the yellow triangle by 90
degrees. This means that this median in the figure at the left hand side must
have been perpendicular to AC_{1}.

In my paper
Friendship
among triangle centers I have shown this in a more indirect and difficult
way.

#### Orthodiagonal quadrangle

In an orthodiagonal quadrangle (the diagonals are perpendicular to each other)
we can use this knowledge in the
following way. Let the figures speak for itself.

So the two sums of the areas of the squares
attached to pairs of opposite sides are equal.

When one side of this figure
shrinks to a point, the figure shows a right triangle and this theorem is the
Pythagorean theorem.

#### The Pythagorean theorem in another way

There is a second way to derive the Pythagorean theorem from this figure,
without a disappearing side. For this we get back to the left hand side of the
first figure. We note that ABC_{2}C_{1} is orthodiagonal as AC_{1}
rotated by 90 degrees about M_{c} becomes BC_{2}.

So the red squares and blue squares have
equal area sums. This shows that the sum of area of the squares APBM_{c}
and C_{1}M_{c}C_{2}Q, which are each half the area of a
blue square, is the mean of the areas of the red squares.

When the angles at M_{c} are right, the Pythagorean theorem follows.

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