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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 McR 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 AC1.
In my paper Friendship among triangle centers I have shown this in a more indirect and difficult way.
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.
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 ABC2C1 is orthodiagonal as AC1 rotated by 90 degrees about Mc becomes BC2.
So the red squares and blue squares have equal area sums. This shows that the sum of area of the squares APBMc and C1McC2Q, which are each half the area of a blue square, is the mean of the areas of the red squares.
When the angles at Mc are right, the Pythagorean theorem follows.
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