SSS Congruence Theorem: Dynamic Proof
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This example demonstrates the SSS Congruence Theorem, which states that if two triangles have all three sides equal, then the triangles are congruent. Key mathematical concepts include triangle congruence and dynamic proof. Using GeoGebra commands such as Circle, Intersect, Segment, and Polygon, the construction process is dynamically shown. Suitable for middle school and above students to learn geometric proofs.
GeoGebra Commands & Parameters
Circle(A, a) → d_1Intersect(d_1, e) → B, CCircle(C, b) → fSegment(A, C) → gCircle(A, c) → kIntersect(k, f) → F, G, ,Segment(C, F) → hSegment(F, A) → iPolygon(A, C, F) → poly1, f_1, a_1, c_1Circle(D, i) → pPoint(p) → ESegment(D, E) → nCircle(D, g) → qCircle(E, h) → rIntersect(q, r, 2) → HSegment(D, H) → sSegment(H, E) → tLine(A, n) → g_1Point(g_1) → IAngle(I, A, F) → αDilate(A, If[0 < o ≤ 1, 1 - o, 0], D) → A'Vector(A, A') → uTranslate(F, u) → F'Translate(A, u) → A'_1Segment(F', A'_1) → i'Rotate(F', If[1 < o ≤ 2, ((-(o - 1)) * α), (-α)], D) → F''Rotate(A'_1, If[1 < o ≤ 2, ((-(o - 1)) * α), (-α)], D) → JSegment(F'', J) → i''Dilate(A, If[2 < o ≤ 3, 3 - o, 0], D) → A'_2Vector(A, A'_2) → vTranslate(A, v) → A'_3Translate(C, v) → C'Segment(A'_3, C') → g'Line(D, g) → h_1Point(h_1) → KAngle(H, D, K) → βRotate(A'_3, If[3 < o ≤ 4, ((-(o - 3)) * β), (-β)], D) → LRotate(C', If[3 < o ≤ 4, ((-(o - 3)) * β), (-β)], D) → C''Segment(L, C'') → g''Line(C, t) → i_1Point(i_1) → MAngle(M, C, F) → γDilate(C, If[4 < o ≤ 5, 5 - o, 0], C'') → C'_1Vector(C, C'_1) → wTranslate(C, w) → C'_2Translate(F, w) → F'_1Segment(C'_2, F'_1) → h'Rotate(C'_2, If[5 < o ≤ 6, ((-(o - 5)) * γ), (-γ)], C'') → NRotate(F'_1, If[5 < o ≤ 6, ((-(o - 5)) * γ), (-γ)], C'') → OSegment(N, O) → h''Line(F, C) → j_1Mirror(A, j_1) → A'_6Mirror(C, j_1) → C'_4Mirror(F, j_1) → F'_3Polygon(A'_6, C'_4, F'_3) → poly1'_1, f'_3, a'_6, c'_4Point(j_1) → SPoint(j_1) → TRay(C'_4, T) → k_1Ray(F'_3, S) → l_1Segment(A'_6, F'_3) → m_1Segment(A'_6, C'_4) → n_1Dilate(F, If[0 < d ≤ 1, 1 - d, 0], E) → F'_2Vector(F, F'_2) → u_1Translate(A'_6, u_1) → PTranslate(C'_4, u_1) → QTranslate(F'_3, u_1) → RPolygon(P, Q, R) → poly2, r_1, p_1, q_1Translate(A'_6, u_1) → UTranslate(C'_4, u_1) → VSegment(U, V) → s_1Translate(A'_6, u_1) → WTranslate(F'_3, u_1) → ZSegment(W, Z) → t_1Translate(C, u_1) → C'_3Translate(F, u_1) → F'_4Segment(C'_3, F'_4) → h'_1Translate(A'_6, u_1) → A_1Translate(F'_3, u_1) → B_1Segment(A_1, B_1) → e_1Line(E, k_1) → f_2Point(f_2) → C_1Angle(C'', E, C_1) → δRotate(P, If[1 < d ≤ 2, ((-(d - 1)) * δ), (-δ)], E) → P'Rotate(Q, If[1 < d ≤ 2, ((-(d - 1)) * δ), (-δ)], E) → Q'Rotate(R, If[1 < d ≤ 2, ((-(d - 1)) * δ), (-δ)], E) → R'Polygon(P', Q', R') → poly2', r', p', q'Rotate(A_1, If[1 < d ≤ 2, ((-(d - 1)) * δ), (-δ)], E) → D_1Rotate(B_1, If[1 < d ≤ 2, ((-(d - 1)) * δ), (-δ)], E) → E_1Segment(D_1, E_1) → g_2Rotate(U, If[1 < d ≤ 2, ((-(d - 1)) * δ), (-δ)], E) → U'Rotate(V, If[1 < d ≤ 2, ((-(d - 1)) * δ), (-δ)], E) → V'Segment(U', V') → h_2Rotate(C'_3, If[1 < d ≤ 2, ((-(d - 1)) * δ), (-δ)], E) → F_1Rotate(F'_4, If[1 < d ≤ 2, ((-(d - 1)) * δ), (-δ)], E) → G_1Segment(F_1, G_1) → i_2Polygon(D, C'', E) → poly3, e_2, d_2, c''Dilate(D, If[2.5 < d ≤ 3.5, 3.5 - d, 0], D_1) → D'Segment(D, D') → j_2Segment(E, C'') → k_2Min(c, k_2 / 2) → o_1Circle(D, (0.3 * o_1)) → c_2Intersect(c_2, i'') → H_1,Intersect(c_2, j_2) → I_1,CircleSector(D, I_1, H_1) → p_2Intersect(j_2, k_2) → J_1Polygon(D, D_1, E) → poly4, e_3, d_3, d_4Dilate(D, If[3.5 < d ≤ 4.5, 4.5 - d, 0], D_1) → D'_1Vector(D, D'_1) → v_1Translate(p_2, v_1) → q_2Angle(E, D_1, D) → εRotate(q_2, If[4.5 < d ≤ 5.5, ((d - 4.5) * (π - ε)), π - ε], D_1) → r_2Intersect(c_2, g'') → K_1, L_1CircleSector(D, K_1, I_1) → s_2Polygon(D, C'', D_1) → poly5, d_5, d_6, c''_1Dilate(D, If[6 < d ≤ 7, 7 - d, 0], D_1) → D'_2Vector(D, D'_2) → w_1Translate(s_2, w_1) → t_2Angle(D, D_1, C'') → ζRotate(t_2, If[7 < d ≤ 8, ((-(d - 7)) * (π - ζ)), (-(π - ζ))], D_1) → c_3Dilate(E, If[8.5 < d ≤ 9.5, 9.5 - d, 0], C'') → E'Segment(E, E') → l_2Segment(E, C'') → m_2Polygon(A, C, F) → poly6, f_3, a_2, c_4Polygon(D, C'', E) → poly7, e_4, d_7, c''_2
AI Drawing Prompt Reference
Draw a dynamic proof of the SSS Congruence Theorem, showing that two triangles are congruent if all three sides are equal. Start by drawing two circles to determine the vertices of the triangles, then connect the sides. Add auxiliary circles, segments, and marked angles. Use transformations like dilation, translation, and rotation to dynamically demonstrate the congruence of the triangles, including the transformation of corresponding sides and angles.
The prompt above was compiled from the original drawing commands and is for reference only.