Dynamic Proof of SAS Triangle Congruence Theorem by Aaron Dankman
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Triangle Congruence Book 1
This example dynamically demonstrates the proof of the SAS Triangle Congruence Theorem. It involves core mathematical concepts such as triangle congruence criteria, angles, and segment lengths. The construction utilizes GeoGebra commands/objects like Circle, Point, Segment, Rotate, Angle, Ray, Intersect, Polygon, CircleSector, Mirror, Line, Dilate, Vector, and Translate. It is suitable for high school mathematics teaching or geometry proof learning.
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. Start with a triangle where two sides have fixed lengths and the included angle is fixed. Construct two circles centered at a vertex, place points on them to form the triangle, and show the angle. Then reflect the entire triangle across one of its sides to create a congruent triangle, displaying corresponding sides and angles. Next, construct another congruent triangle at a different location using rotation and circle intersections. Finally, add a helper line with points on it, create a translation vector, and translate part of the first triangle to show equal corresponding sides. The figure includes triangles, circles, rays, segments, angle marks, and reflected images.
The prompt above was compiled from the original drawing commands and is for reference only.