Then on each parallel the transversals cut two segments in the same proportion: AY

The inverse is also true. If three transversals a, b and c cut on parallels e

To show, for example that BY

From this follows AY

The inverse is proved by reversing the argument and applying the inverse of Thales theorem.

In fact, line c is the median of all triangles OEE', ODD', ... . We see this by examining one of this triangles and applying the previous theorem. Preferably we draw a parallel from A* and show that A*A

For this subtract from triangles AA'B and AA'B' their common part AA'A* and see that area(AA*B) = area(A'A*B'). Then, area(AA*B) = dA*A

A'B/A'C=B'C/B'A=C'A/C'B=k και A''B/A''C=B''C/B''A=C''A/C''B=k' (*).

Join the dividing points with the opposite sides and construct through the intersections of the various lines the various triangles, quadrangles, pentagons and the central hexagon of the figure shown. Show that:

[1] The area of the resulting polygons remain unchanged if a vertex of ABC is translated along the parallel to the opposite side.

[2] Triangles with the same color have equal areas.

[3] The ratio of areas of each of the triangles, quadrangles, pentagons and central hexagon to the area of triangle ABC is independent of the shape of ABC.

In all these variations of the shape of ABC we assume that the ratios defining the dividing points {A',A'',B',B'',C',C''} remain constant (i.e. (*) remains valid all the time).

From [1] follows that for the proof of the other two properties we can restrict ourselves to an

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