Notes on The Pythagorean Theorem

First consider Euclid's statement and proof of Proposition 47.

In right-angled triangles the square on the side opposite the right angle equals the sum of the squares on the sides containing the right angle.

Note that Euclid's treatment in its statement or its "proof" never refers the traditional equation, a²+b²=c².
In one alternative proof for this theorem illustrated in the java sketch below, we consider 4 congruent right triangles and 2 squares and then the same 4 triangles and the square on the side of the hypotenuse arranged inside of a square with side "a+b" . Can you explain how this sketch justifies the theorem?

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Shear the squares on the legs by dragging point P, then point Q, to the line. Shearing does not affect a polygon's area.
Shear the square on the hypotenuse by dragging point R to fill the right angle.
The resulting shapes are congruent.
Therefore, the sum of the squares on the sides equals the square on the hypotenuse.

How could we justify identifying "equal" objects (congruent figures)?
How do the objects fit together?
How do movements effect the shapes of objects?

[Side Trip] Moving line segments:

Consider Euclid's Proposition 1 and Proposition 2.

These propositions demonstrate that Euclid did not treat moving a line segment as an essential property worthy of being at the foundations as an axiom. However, this is a fundamental tool for all of geometry.
Note that in the proofs of propositions 1 and 2 certain points of intersection of circles are presumed to exist without reference to any of the postulates. These presumptions were left implicit for hundreds of years, but were cleared up in the 19th century when careful attention was given again to the axioms as a whole system. This presumption is sometimes described as a postulate of "continuity" for lines and circles.

[An example of a geometry where circles do not intersect is given by using the rational coordinate plane. Points correspond to ordered pairs of rational numbers. then the circle with center (0,0) and radius 1 and the circle with center (1,0) also radius 1 meet in the ordinary plane at the points with coordinates (1/2, sqrt(3)/2) and (1/2, -sqrt(3)/2) . Since sqrt(3)/2 is not a rational number, this ordered pair does not correspond to a point in the rational coordinate plane, so the two circles do not have a point of intersection.

Another example of a point not in the rational coordinate plane is the point (sqrt(2),0). This point can be constructed in the ordinary plane with straight edge and compass using the circle with center (0,0) and radius determined by the points (0,0) and (1,1). This circle will meet the X-coordinate axis at the point (sqrt(2),0) ]