Pythagorean Theorem via Similar Parallelograms (Dynamic & Modifiable)

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2019-06-04 21:58

① This interactive GeoGebra example demonstrates a geometric proof of the Pythagorean theorem by constructing similar parallelograms on the three sides of a right triangle, showing that the sum of the areas of the two smaller parallelograms equals the area of the larger one. ② Core mathematical concepts include the Pythagorean theorem, similarity, area relationships, parallelogram properties, and dilation transformations. ③ Construction highlights: using GeoGebra commands such as Semicircle, Point, Polygon, Rotate, Ray, Circle, Intersect, Line, Midpoint, OrthogonalLine, Dilate, Vector, Translate, Segment, Angle, and Distance, the applet builds a right triangle and similar parallelograms on each side, with dynamic sliders or draggable points to modify the shape and observe area changes in real time. ④ Suitable for middle and high school geometry classes, dynamic proof demonstrations, and exploratory learning of area-based proofs and similarity.

GeoGebra Commands & Parameters

Semicircle(A, B) → c
Point(c) → C
Polygon(A, C, B) → t1, b, a, c_1
Rotate(B, (d * π / 180), C) → B'
Ray(C, B') → f
Rotate(C, (d * π / 180), A) → C'
Ray(A, C') → g
Rotate(A, (d * π / 180), B) → A'
Ray(B, A') → h
Point(Segment[C, B']) → D
Circle(A, (e * Distance[A, C'])) → k
Circle(B, (e * Distance[B, A'])) → p
Intersect(k, g, 1) → E
Intersect(p, h, 1) → F
Line(D, a) → i
Line(E, b) → j
Line(F, c_1) → l
Line(B, f) → m
Line(A, h) → n
Line(C, g) → q
Intersect(j, q) → G
Intersect(i, m) → H
Intersect(l, n) → I
Polygon(A, E, G, C) → q1, a_1, e_1, g_1, c_2
Polygon(C, D, H, B) → q2, c_3, d_1, h_1, b_1
Polygon(B, F, I, A) → q3, b_2, f_1, i_1, a_2
Intersect(f_1, i_1) → J
Midpoint(I, B) → K
Midpoint(A, G) → L
Midpoint(C, H) → M
OrthogonalLine(C, a_2) → r
Ray(B, H) → s
Intersect(r, a_2) → N
OrthogonalLine(D, b_1) → t
Intersect(b_1, t) → O
Line(D, H) → j_1
Line(C, B) → k_1
Circle(B, i_1) → p_1
Intersect(p_1, j_1, 2) → P
Dilate(H, If[0 < u ≤ 1, 1 - u, 0], P) → H'
Vector(H, H') → v
Translate(D, v) → D'
Polygon(C, B, H', D') → q4, c_4, b_3, h', d'
Line(C, D') → l_1
Line(I, F) → m_1
Line(N, h) → n_1
Intersect(m_1, n_1) → Q
Segment(N, Q) → r_1
Segment(B, P) → s_1
Polygon(Q, F, B, N) → q5, q_2, f_2, b_4, n_2
Circle(B, b_4) → q_1
Intersect(q_1, l_1, 1) → R
Dilate(C, If[2 < u ≤ 3, 3 - u, 0], R) → C'_1
Vector(C, C'_1) → w
Translate(D', w) → D''
Polygon(B, P, D'', C'_1) → q6, b_5, p_2, d'', c'_1
Midpoint(Q, B) → S
Angle(R, B, N) → α
Rotate(B, If[4 < u ≤ 5, ((u - 4) * α), α], B) → B'_1
Rotate(P, If[4 < u ≤ 5, ((u - 4) * α), α], B) → P'
Rotate(D'', If[4 < u ≤ 5, ((u - 4) * α), α], B) → D'''
Rotate(C'_1, If[4 < u ≤ 5, ((u - 4) * α), α], B) → T
Polygon(B'_1, P', D''', T) → q6', b'_1, p', d''', t_1
Dilate(N, If[5 < u ≤ 6, 6 - u, 0], Q) → N'
Vector(N, N') → u_1
Translate(B'_1, u_1) → U
Translate(P', u_1) → P''
Translate(D''', u_1) → V
Translate(T, u_1) → T'
Polygon(U, P'', V, T') → q6'', u_2, p'', v_1, t'
Line(A, C) → g_2
Line(E, G) → h_2
Circle(C, Distance[A, J]) → d_2
Intersect(d_2, h_2, 2) → W
Dilate(G, If[7 < u ≤ 8, 8 - u, 0], W) → G'
Vector(G, G') → w_1
Translate(E, w_1) → E'
Polygon(A, C, G', E') → q7, a_3, c_5, g', e'
Circle(W, Distance[A, V]) → e_2
Distance(C, W) → distanceCW
Distance(A, I) → distanceAI
Line(C, W) → i_2
Circle(W, Distance[A, V]) → k_2
Intersect(k_2, j_1, 1) → A_1
Line(A, i_2) → j_2
Intersect(j_2, h_2) → B_1
Intersect(k_2, j_2, 1) → C_1
Dilate(B_1, If[9 < u ≤ 10, 10 - u, 0], C_1) → D_1
Vector(B_1, D_1) → v_2
Translate(A, v_2) → A'_1
Polygon(C, A'_1, D_1, W) → q8, l_2, m_2, r_2, s_2
Dilate(A'_1, If[11 < u ≤ 12, 12 - u, 0], A) → E_1
Vector(A'_1, E_1) → w_2
Translate(C, w_2) → C'_2
Translate(A'_1, w_2) → F_1
Translate(D_1, w_2) → G_1
Translate(W, w_2) → W'
Polygon(C'_2, F_1, G_1, W') → q8', c'_2, t_2, f_3, w'
Line(E_1, C'_2) → g_3
Line(E_1, b_4) → h_3
Point(h_3) → H_1
Angle(H_1, E_1, C'_2) → β
Distance(E_1, C'_2) → distanceE1C'2
Distance(A, N) → distanceAN
Rotate(C'_2, If[12 < u < 13, ((-(u - 12)) * β), (-β)], F_1) → I_1
Rotate(F_1, If[12 < u < 13, ((-(u - 12)) * β), (-β)], F_1) → J_1
Rotate(G_1, If[12 < u < 13, ((-(u - 12)) * β), (-β)], F_1) → K_1
Rotate(W', If[12 < u < 13, ((-(u - 12)) * β), (-β)], F_1) → W''
Polygon(I_1, J_1, K_1, W'') → q8'', i_3, j_3, k_3, w''
Dilate(A, If[13 < u ≤ 14, 14 - u, 0], J) → A'_2
Vector(A, A'_2) → u_3
Translate(I_1, u_3) → L_1
Translate(J_1, u_3) → M_1
Translate(K_1, u_3) → N_1
Translate(W'', u_3) → W'''
Polygon(L_1, M_1, N_1, W''') → q8''', l_3, m_3, n_3, w'''
Ray(E, G) → p_3
Ray(D, H) → q_3
Ray(C, B) → r_3
Ray(A, C) → s_3
Ray(C, D') → t_3
Ray(B, H') → d_3
Segment(A, E) → e_3
Ray(A, B_1) → f_4
Ray(C, W) → g_4
Segment(C, D) → h_4
Angle(A, C, B) → γ
Polygon(A, C, B) → t2, b_6, a_4, c_6
Segment(A, I) → i_4
Segment(I, F) → j_4
Segment(F, B) → k_4
Segment(B, A) → l_4
Segment(C, B) → m_4
Segment(C, A) → n_4

AI Drawing Prompt Reference

Draw a right triangle with the hypotenuse as the diameter of a semicircle, and place a point on the arc to form the right angle vertex. Construct three similar parallelograms outward on each side of the triangle, such that the sum of the areas of the two smaller ones equals the area of the largest. Use rotating rays and circles to determine the parallelogram vertices, and include draggable points or sliders to adjust angles and proportions dynamically, showing area changes in real time. Finally, mark the right angle and key segments to illustrate the area relationship.

The prompt above was compiled from the original drawing commands and is for reference only.