Pythagorean Theorem VI (visual proof; Euclid-inspired proof; 4K)

Start with a right triangle with legs of length a, b, and a hypotenuse of length c. The square drawn using the side of length a has an area of a squared. The square drawn using the side of length b has an area of b squared. And the square drawn using the hypotenuse of length c has an area of c squared. Now draw lines extending the sides of the two smaller squares like this. These lines intersect and are perpendicular by construction. Note that they also enclose an a by b rectangle. We can transform the a squared area along the line creating a parallelogram. This parallelogram has the same width and height as the original square so its area is still a squared. We can do the same thing to the b squared area creating a parallelogram with the same width and height as the original square so its area is still b squared. Both of these new parallelograms have a side of length c because we shifted the small square b units along its line and the medium square a units along its line. We can thus shift both parallelograms down c units. And then they hit the bottom of the large square. The extra portion covers the original right triangle and we see that we can shift that right triangle down to fill the uncovered part of the large square. Because the large square is completely covered by the area of the two parallelograms we conclude that a squared plus b squared is equal to c squared. Remembering that a and b are the lengths of two legs of a right triangle with hypotenuse of length c, we have shown that for such a right triangle a squared plus b squared equals c squared. This is the Pythagorean theorem essentially using Euclid's proof.