Triangle Midsegment Theorem Dynamic Illustration
Loading the math board and drawing, please wait…
In collections
Congruence (CO) Cluster
Congruence (Vol 2)
CCSS High School: Geometry (Congruence) Volume 2
CCSS High School: Geometry (Congruence) Volume 2
CCSS High School: Geometry (Congruence) Volume 2
Geometry (Congruence) Volume 2
CCSS High School: Geometry (Congruence) Volume 2
CCSS High School: Geometry (Congruence) Volume 2
CCSS High School: Geometry (Congruence) Volume 2
CCSS High School: Geometry (Congruence) Volume 2
CCSS High School: Geometry (Congruence) Volume 2
CCSS High School: Geometry (Congruence) Volume 2
CCSS High School: Geometry (Congruence) Volume 2
Gail LeBlancCCSS High School: Geometry (Congruence) Volume 2
CCSS High School: Geometry (Congruence) Volume 2
Triangle Theorems
This example dynamically illustrates the Triangle Midsegment Theorem without words or numbers. It includes a definition prompt and a guiding proof question. Commands used include Segment, Line, Slope, PointIn, Min, Midpoint, Circle, Intersect, Angle, Rotate, Polygon, Dilate, Vector, CircleSector, Translate. Suitable for middle school and above students learning geometric construction.
GeoGebra Commands & Parameters
Segment(A, B) → gLine(A, B) → fSlope(f) → mPointIn(a) → CSegment(C, A) → iSegment(B, C) → hMin(list1) → dMidpoint(C, A) → FCircle(F, (0.15 * d)) → e_1Midpoint(B, C) → ELine(F, E) → i_1Intersect(e_1, i_1) → O, PAngle(P, F, C) → βRotate(P, If[9 < l ≤ 10, ((l - 9) * β), β], F) → P'_2Polygon(A, B, C) → poly1, c, a_1, b_1Midpoint(A, B) → DCircle(F, (0.15 * d)) → eSegment(F, E) → jIntersect(e, i) → G, HIntersect(e, j, 1) → IDilate(A, If[0 < l ≤ 1, 1 - l, 0], D) → A'Segment(A, A') → nDilate(D, If[1 < l ≤ 2, 2 - l, 0], B) → D'Segment(D, D') → pRotate(A, If[2 < l ≤ 3, ((-(l - 2)) * π), (-π)], D) → A'_1Rotate(A', If[2 < l ≤ 3, ((l - 2) * π), π], D) → A''Segment(A'_1, A'') → n'Dilate(B, If[3 < l ≤ 4, 4 - l, 0], E) → B'Segment(B, B') → qDilate(E, If[4 < l ≤ 5, 5 - l, 0], C) → E'Segment(E, E') → rRotate(B, If[5 < l ≤ 6, ((-(l - 5)) * π), (-π)], E) → B'_1Rotate(B', If[5 < l ≤ 6, ((l - 5) * π), π], E) → B''Segment(B'_1, B'') → q'Dilate(C, If[6 < l ≤ 7, 7 - l, 0], F) → C'Segment(C, C') → sDilate(F, If[7 < l ≤ 8, 8 - l, 0], A) → F'Segment(F, F') → tRotate(C, If[8 < l ≤ 9, ((-(l - 8)) * π), (-π)], F) → C'_1Rotate(C', If[8 < l ≤ 9, ((l - 8) * π), π], F) → C''Segment(C'_1, C'') → s'Dilate(F, If[9 < l ≤ 10, 10 - l, 0], E) → F'_1Segment(F, F'_1) → f_1Dilate(F, If[10 < l ≤ 11, 11 - l, 0], A) → F'_2Vector(F, F'_2) → uCircleSector(F, P, P'_2) → kTranslate(k, u) → k'Dilate(E, If[11 < l ≤ 12, 12 - l, 0], D) → E'_1Vector(E, E'_1) → vTranslate(F, v) → F'_3Translate(F'_1, v) → JSegment(F'_3, J) → g_1Dilate(F, If[12 < l ≤ 13, 13 - l, 0], D) → F'_4Vector(F, F'_4) → wTranslate(F, w) → F'_5Translate(F'_1, w) → KSegment(F'_5, K) → h_1Angle(C, B, A) → αIntersect(e, j) → L,Intersect(e, i, 1) → MIntersect(i, j) → NCircleSector(N, M, L) → c_1Translate(c_1, u) → d_1Rotate(P, If[9 < l ≤ 10, l - 9]) → P'Rotate(P, If[9 < l ≤ 10, ((l - 9) * β), β]) → P'_1
AI Drawing Prompt Reference
Create a dynamic geometric illustration of the triangle midsegment theorem. Start with an arbitrary triangle, mark the midpoints of all three sides, and connect two midpoints to form a midsegment. Show animated steps where vertices slide along sides, segments are translated or rotated, and small circles highlight midpoints. Use arcs or arrows to indicate rotation or translation directions. The final visual should demonstrate that the midsegment is parallel to the third side and half its length, without any text or numbers.
The prompt above was compiled from the original drawing commands and is for reference only.