Visual Proof of Triangle Area Formula: Dissection into an Equivalent Parallelogr
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Euclidean Geometry
This GeoGebra example presents a visual dissection proof of the triangle area formula A = (b · h) / 2. It shows how a triangle can be cut and rearranged—using midpoint, perpendicular, intersection, segment, line, and rotate operations—to form an equivalent parallelogram with the same base and height, proving that the triangle’s area is half the parallelogram’s area. Key mathematical concepts include triangle area, parallelogram area, equidecomposability, altitude, base-height relationship, and geometric construction. The construction relies on GeoGebra commands Polygon, OrthogonalLine, Intersect, Midpoint, Segment, Line, and Rotate, involving objects such as vertices A, B, C, altitude foot H, and auxiliary points D, E, F. It is ideal for middle school geometry lessons, area formula derivation, teacher preparation, and student-centered exploration of transformations and area invariance.
GeoGebra Commands & Parameters
Polygon(A, B, C) → t1, c, a, bOrthogonalLine(C, c) → fIntersect(f, c) → HMidpoint(C, H) → DSegment(C, D) → gSegment(D, H) → hLine(D, c) → iIntersect(i, a) → EIntersect(i, b) → FPolygon(C, F, E) → t2, e, c_1, f_1Polygon(A, B, E, F) → q1, a_1, b_1, e_1, f_2Rotate(C, (-α), E) → C'_1Rotate(F, (-α), E) → F'_1Rotate(E, (-α), E) → E'_1Polygon(C'_1, F'_1, E'_1) → t2'_1, e'_1, c'_1, f'_1Rotate(D, (-α), E) → D'Segment(C'_1, D') → j
AI Drawing Prompt Reference
Draw a triangle ABC, construct a perpendicular line from C to AB intersecting at H, find the midpoint D of CH, connect CD and DH, draw a line through D parallel to AB intersecting AC at E and BC at F, form triangle CEF, then rotate triangle CEF counterclockwise around E to form a new triangle C'EF', and connect C' with the rotated point D'. This will create a parallelogram and two triangles, demonstrating the proof of the triangle area formula A=(b·h)/2.
The prompt above was compiled from the original drawing commands and is for reference only.