From a point D inside the triangle t=(ABC) draw parallels to the sides. Transfer isometrically (without reflections) the resulting triangles (DEF), (DHG) and (DIJ) to (D'E'F'), (D'H'G') and (D'I'J'), correspondingly, so that their vertices coincide at D', as they did in D. Build the parallelograms from their sides: (A'I'D'G'), (B'E'D'G') and (C'H'D'F'). Then the triangle t' = (A'B'C') is always similar to t.

In the figure above, points K, L, M and D' are freely movable control-points.

By changing their places you access all possible placements of the triangles (D'E'F'), (D'H'G') and (D'I'J'), as required by the theorem. An exact (rigid) copy of triangle t' = (A'B'C') is the triangle t''=(A''B''C''). The angle-measures of the later are displayed. By changing the positions of the control points you see that the angles remain constant.

The theorem is also known as the [Fundamental theorem of 3-Bar Motion]. In this formulation the 15 segments composing the figure are considered as rods (fixed length), connected at the 10 swivel-joints (points A, B, C, D, E, ...).

A discussion and proof of the theorem is contained in [Ross Honsberger's, In Polya's Footsteps, The Mathematical Assoc. of America, 1997, p. 129].

Another example, that moves linked polygons can be found in the file: Dudeney.html .

G. B. Halsted,

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