MathPlotAI Logo MathPlotAI

Pythagorean Theorem Proof Without Words

Pythagorean Theorem Proof Without Words
Loading the math board and drawing, please wait…
2021-02-28 09:18
TrianglePolygonAngleGeometric Construction
① This example visually demonstrates the Pythagorean theorem, showing that the square of the hypotenuse of a right triangle equals the sum of the squares of the other two sides. ② It involves the Pythagorean theorem, properties of right triangles, area relationships, similarity, and geometric constructions using rotations, reflections, and congruences. ③ The GeoGebra construction uses commands such as Polygon, Point, Rotate, Line, Angle, OrthogonalLine, AngularBisector, Intersect, LineBisector, Circle, Midpoint, and conditional statements to create the figure and animate the proof. ④ Designed for high school geometry or middle school mathematics, suitable for classroom demonstrations, problem solving, and student exploration.

GeoGebra Commands & Parameters

  1. Polygon(A, B, 4) → poly1, f, g, h, i, C, D
  2. Point(Segment[C, (0, 1)]) → E
  3. Rotate(F, π / 2) → F'
  4. Rotate(E, π / 2) → E'
  5. Line(E', E) → l_1
  6. Line(E', F') → l
  7. Angle(xAxis, l) → α
  8. Line(C, l) → n
  9. OrthogonalLine(C, n) → h_1
  10. AngularBisector(h_1, n) → i_1, j_1
  11. Line(A, D) → p
  12. Intersect(n, p) → L
  13. LineBisector(L, C) → t
  14. Intersect(t, j_1) → T
  15. Line(D, B) → r
  16. Circle(G, Segment[T, C]) → c_1
  17. Intersect(c_1, r) → U, V
  18. Rotate(r, γ, U) → r'
  19. LineBisector(L, D) → q
  20. Intersect(q, r) → N
  21. Polygon(E', G, E, D) → q1, e', g_1, e, d
  22. Polygon(E', A, F, G) → q2, j, a, f_1, g_2
  23. Polygon(F, B, F', G) → q3, f_2, b, f', g_3
  24. Polygon(G, F', C, E) → q4, g_4, k, c, e_1
  25. Line(F, E) → m
  26. Midpoint(L, D) → M
  27. Midpoint(L, C) → S
  28. Line(A, C) → k_1
  29. Intersect(c_1, k_1) → W, Z
  30. Intersect(l, e') → H
  31. Intersect(r', l_1) → K
  32. Intersect(l, e') → I
  33. If(0 < o ≤ 1, Polygon[Rotate[E', ((o - 0) * (π + α)), K], Rotate[G, ((o - 0) * (π + α)), K], Rotate[E, ((o - 0) * (π + α)), K], Rotate[D, ((o - 0) * (π + α)), K]], If[o > 1, Polygon[Rotate[E', π + α, K], Rotate[G, π + α, K], Rotate[E, π + α, K], Rotate[D, π + α, K]]]) → q6
  34. Midpoint(G, E) → J
  35. Circle(G, Segment[N, L]) → s
  36. Rotate(K, π / 2, (0, 0)) → K'
  37. Rotate(K, π, (0, 0)) → K'_1
  38. Rotate(K, (3 * π / 2), (0, 0)) → K'_2
  39. If(2 < o ≤ 3, Polygon[Rotate[E', ((o - 2) * (π + α)), K'], Rotate[A, ((o - 2) * (π + α)), K'], Rotate[F, ((o - 2) * (π + α)), K'], Rotate[G, ((o - 2) * (π + α)), K']], If[o > 3, Polygon[Rotate[E', π + α, K'], Rotate[A, π + α, K'], Rotate[F, π + α, K'], Rotate[G, π + α, K']]]) → q5
  40. If(4 < o ≤ 5, Polygon[Rotate[F, ((o - 4) * (π + α)), K'_1], Rotate[B, ((o - 4) * (π + α)), K'_1], Rotate[F', ((o - 4) * (π + α)), K'_1], Rotate[G, ((o - 4) * (π + α)), K'_1]], If[o > 5, Polygon[Rotate[F, π + α, K'_1], Rotate[B, π + α, K'_1], Rotate[F', π + α, K'_1], Rotate[G, π + α, K'_1]]]) → q7
  41. If(6 < o ≤ 7, Polygon[Rotate[G, ((o - 6) * (π + α)), K'_2], Rotate[F', ((o - 6) * (π + α)), K'_2], Rotate[C, ((o - 6) * (π + α)), K'_2], Rotate[E, ((o - 6) * (π + α)), K'_2]], If[o > 7, Polygon[Rotate[G, π + α, K'_2], Rotate[F', π + α, K'_2], Rotate[C, π + α, K'_2], Rotate[E, π + α, K'_2]]]) → q8
  42. Angle(K, U, G) → β
  43. Intersect(r', l_1) → O

AI Drawing Prompt Reference

Draw a right triangle and construct squares on each of its three sides. Then dissect the two squares on the legs into four congruent pieces each, and using rotations and reflections move those pieces step by step into the square on the hypotenuse, animating the rearrangement to give a wordless visual proof that the area of the hypotenuse square equals the sum of the areas of the leg squares.
The prompt above was compiled from the original drawing commands and is for reference only.