Dynamic Proof of SAS Triangle Congruence Theorem by Luis Alberto Peralta Amaya
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This example dynamically demonstrates the proof of the SAS Triangle Congruence Theorem. It involves core mathematical concepts such as triangle congruence, geometric transformations, rotation, and reflection. The construction utilizes GeoGebra commands and objects like Circle, Point, Segment, Rotate, Angle, Ray, Intersect, Min, Polygon, CircleSector, Mirror, Line, Dilate, Vector, and Translate. It is suitable for high school mathematics education and understanding the dynamic proof of geometric theorems.
GeoGebra Commands & Parameters
Circle(A, a) → cPoint(c) → BSegment(A, B) → fCircle(A, b) → dRotate(B, α, A) → B'Angle(B, A, B') → βRay(A, B') → gIntersect(d, g, 1) → CSegment(B, C) → hSegment(A, C) → iMin(f, i) → eCircle(A, (0.25 * e)) → kIntersect(k, f) → D,Intersect(k, i) → E,Polygon(A, B, C) → poly1, c_1, a_1, b_1CircleSector(A, D, E) → pMirror(A, h) → A'Mirror(B, h) → B'_1Mirror(C, h) → C'Polygon(A', B'_1, C') → poly1', c', a', b'_1Mirror(p, h) → p'Mirror(A, h) → A'_1Mirror(B, h) → B'_2Segment(A'_1, B'_2) → f'Mirror(A, h) → A'_2Mirror(C, h) → C'_1Segment(A'_2, C'_1) → i'Line(C, B) → jPoint(j) → GPoint(j) → HRay(B, H) → lRay(C, G) → mCircle(F, a) → qPoint(q) → ISegment(F, I) → nRotate(I, α, F) → I'Angle(I, F, I') → γCircle(F, b) → rRay(F, I') → sIntersect(r, s, 1) → JSegment(I, J) → tSegment(F, J) → f_1Dilate(B, If[0 < u ≤ 1, 1 - u, 0], I) → B'_3Vector(B, B'_3) → vTranslate(A, v) → A'_3Translate(B, v) → B'_4Segment(A'_3, B'_4) → f'_1Line(B, n) → g_1Point(g_1) → KAngle(K, B, A) → δRotate(A'_3, If[1 < u ≤ 2, ((-(u - 1)) * δ), (-δ)], I) → LRotate(B'_4, If[1 < u ≤ 2, ((-(u - 1)) * δ), (-δ)], I) → MSegment(L, M) → h_1Dilate(A, If[2 < u ≤ 3, 3 - u, 0], F) → A'_4Vector(A, A'_4) → wTranslate(A, w) → A'_5Translate(C, w) → C'_2Segment(A'_5, C'_2) → i'_1Line(A, f_1) → i_1Point(i_1) → NAngle(N, A, C) → εRotate(A'_5, If[3 < u ≤ 4, ((-(u - 3)) * ε), (-ε)], F) → ORotate(C'_2, If[3 < u ≤ 4, ((-(u - 3)) * ε), (-ε)], F) → PSegment(O, P) → j_1Translate(p, w) → p'_1Rotate(p'_1, If[3 < u ≤ 4, ((-(u - 3)) * ε), (-ε)], F) → d_1Dilate(B, If[4 < u ≤ 5, 5 - u, 0], I) → B'_5Vector(B, B'_5) → u_1Translate(A, u_1) → A'_6Translate(B, u_1) → B'_6Translate(C, u_1) → C'_3Polygon(A'_6, B'_6, C'_3) → poly1'_1, c'_3, a'_6, b'_6Line(B, t) → k_1Point(k_1) → QAngle(Q, B, C) → ζRotate(A'_6, If[5 < u ≤ 6, ((-(u - 5)) * ε), (-ε)], I) → RRotate(B'_6, If[5 < u ≤ 6, ((-(u - 5)) * ε), (-ε)], I) → SRotate(C'_3, If[5 < u ≤ 6, ((-(u - 5)) * ε), (-ε)], I) → TPolygon(R, S, T) → poly2, t_1, r_1, s_1Segment(B, A) → l_1Segment(A, C) → m_1Segment(C, B) → n_1CircleSector(A, D, E) → e_1
AI Drawing Prompt Reference
Draw a dynamic proof of the SAS Triangle Congruence Theorem, including two triangles and a circle, showing congruence through rotation and reflection, with a slider controlling the angle, and labeling key points and segments to illustrate the geometric transformation process.
The prompt above was compiled from the original drawing commands and is for reference only.