Thales

Thales theorem

Given are parallel lines e₁, e₂, ... and transversals to them a, b, ... . The parallel lines cut on a, b, ... segments whose proportions are the same. In the figure below AB/BC = A'B'/B'C', BC/CD = B'C'/C'D' etc. ... .
Often we write AB : BC : CD : DE ... = A'B' : B'C' : C'D' : D'E' ... .

The inverse is also true. If lines e₁, e₂, e₃, ... intersect on two transversals a,b proportional segments (i.e. satisfying AB : BC : CD : DE ... = A'B' : B'C' : C'D' : D'E' ...), then e₁, e₂, ... are parallel.

A proof based on a preliminary discussion of the properties of areas of triangles and parallelogramms goes as follows:
The quotient of areas of triangles: area(AB'B)/area(BB'C) = AB/BC. Analogously area(A'BB')/area(B'BC') = A'B'/B'C'. But area(AB'B) = area(A'BB') and area(BB'C) = area(B'BC') etc..
The argument can be reversed to prove the inverse theorem.

An important consequence relating points A*, B*, ... and O are discussed in Thales2.html .

Thales theorem

See Also