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Pythagorean Theorem Puzzle by milanofff

Pythagorean Theorem Puzzle by milanofff
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2012-07-25 23:23
TrianglePolygonCoordinate GeometryGeometric ConstructionSlider Animation
This example demonstrates the Pythagorean theorem, which states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. It involves key mathematical concepts such as triangles, polygons, and geometric construction. Using GeoGebra commands like If, Semicircle, Point, Polygon, etc., it dynamically illustrates the proof of the Pythagorean theorem. Suitable for middle school and above students to learn geometry and algebra.

GeoGebra Commands & Parameters

  1. If(n ≤ 10, 0, If[n ≤ 11, n - 10, 1]) → n11
  2. If(n ≤ 9, 0, If[n ≤ 10, n - 9, 1]) → n10
  3. Semicircle(A, B) → z
  4. Point(z) → C
  5. Polygon(A, B, C) → poly1, c, a, b
  6. Polygon(A, C, 4) → poly3, a_2, c_2, f, g, F, G
  7. Polygon(C, B, 4) → poly4, c_3, b_2, h, i, H, I
  8. OrthogonalLine(I, c) → k
  9. Line(G, H) → j
  10. Intersect(j, k) → K
  11. Polygon(Rotate[C, (n10 * 90)°, I], Rotate[K, (n10 * 90)°, I], Rotate[I, (n10 * 90)°, I]) → poly11, q_1, r_1, s_1
  12. If(n ≤ 4, 0, If[n ≤ 5, n - 4, 1]) → n5
  13. Polygon(B, A, 4) → poly2, b_1, a_1, d, e, D, E
  14. OrthogonalLine(B, c) → l
  15. Intersect(j, l) → J
  16. Angle(A, C, B) → α
  17. OrthogonalLine(F, c) → m
  18. OrthogonalLine(A, c) → p
  19. Intersect(j, p) → L
  20. Intersect(j, m) → M
  21. If(n ≤ 0, 0, If[n ≤ 1, n, 1]) → n1
  22. Translate(J, Vector[n5 * Vector[B, E]]) → J1
  23. Polygon(J1, Translate[J1, Vector[J, B]], Translate[J1, Vector[J, C]]) → poly5, q, j1, r
  24. If(n ≤ 1, 0, If[n ≤ 2, n - 1, 1]) → n2
  25. If(n ≤ 2, 0, If[n ≤ 3, n - 2, 1]) → n3
  26. If(n ≤ 3, 0, If[n ≤ 4, n - 3, 1]) → n4
  27. If(n ≤ 5, 0, If[n ≤ 6, n - 5, 1]) → n6
  28. If(n ≤ 6, 0, If[n ≤ 7, n - 6, 1]) → n7
  29. If(n ≤ 7, 0, If[n ≤ 8, n - 7, 1]) → n8
  30. If(n ≤ 8, 0, If[n ≤ 9, n - 8, 1]) → n9
  31. Polygon(Rotate[B, (n1 * 90)°, B], Rotate[H, (n1 * 90)°, B], Rotate[J, (n1 * 90)°, B]) → poly6, s, t, c_1
  32. Translate(poly6, Vector[n3 * Vector[B, E]]) → poly6'
  33. Polygon(Rotate[A, (n2 * 270)°, A], Rotate[L, (n2 * 270)°, A], Rotate[G, (n2 * 270)°, A]) → poly7, d_1, e_1, f_1
  34. Translate(poly7, Vector[n3 * Vector[A, D]]) → poly7'
  35. Polygon(Translate[A, Vector[n4 * Vector[A, D]]], Translate[C, Vector[n4 * Vector[A, D]]], Translate[L, Vector[n4 * Vector[A, D]]]) → poly8, g_1, h_1, i_1
  36. Polygon(Translate[M, Vector[n6 * Vector[F, B]]], Translate[F, Vector[n6 * Vector[F, B]]], Translate[G, Vector[n6 * Vector[F, B]]]) → poly9, j_1, k_1, l_1
  37. Polygon(Translate[K, Vector[n7 * Vector[I, A]]], Translate[H, Vector[n7 * Vector[I, A]]], Translate[I, Vector[n7 * Vector[I, A]]]) → poly10, m_1, n_1, p_1
  38. Polygon(Rotate[F, (n8 * 270)°, F], Rotate[M, (n8 * 270)°, F], Rotate[C, (n8 * 270)°, F]) → poly12, t_1, d_2, e_2
  39. Translate(poly11, Vector[n11 * Vector[I, A]]) → poly11'
  40. Translate(poly12, Vector[n9 * Vector[F, B]]) → poly12'

AI Drawing Prompt Reference

Please draw a geometric figure that demonstrates the Pythagorean theorem. First, draw a semicircle and an inscribed right triangle within the semicircle. Then, construct squares on each side of the triangle. Next, use perpendicular and parallel lines to assist in proving the Pythagorean theorem. Finally, through rotation and translation operations, dynamically display the proof process of the Pythagorean theorem. The figure includes a semicircle, right triangle, squares, and multiple auxiliary lines.
The prompt above was compiled from the original drawing commands and is for reference only.