I am teaching a geometry course and I am trying to understand two definitions in the textbook ("Geometry with Geometry Explorer" by Michael Hvidsten.)
Definition: The area of a rectangle is its base times its height.
Definition: If two figures can be made equivalent, we will say that they have the same area.
Here we say that two figures can be made equivalent if each can be split into the same finite number of polygons (without loss of generality, triangles) such that corresponding pairs are congruent. Note that "split" does not exactly mean "partitioned" here, because we allow the edges of the triangles to overlap.
To me it seems that the first definition is defining "area" in a special case, and the second definition is defining "has the same area." However, we are clearly meant to infer that if two rectangles "have the same area" in the second sense then they have the same "area" in the first sense. Is this obvious?
Of course one can prove it using analytic methods because triangles are Lebesgue measurable. However, the course takes a synthetic approach to geometry, so it would be better to avoid this. So my question is the following:
Is there a proof in elementary synthetic geometry that two rectangles $R$ and $R'$ with different values for "base times height" (e.g. $1 \times 1$ and $2 \times 1$) cannot be split into finitely many triangles $T_i,\ldots,T_n$ and $T'_1,\ldots,T'_n$ respectively, with $T_i \cong T'_i$ for all $i \le n$?
I think what you want is in Chapter 5 of Hartshorne's book "Geometry: Euclid and Beyond", which covers Hilbert's synthetic theory of area. The theorem you want is a slight rewording of the statement in Andre's post, namely: there is a finitely additive function defined on all figures in the plane that extends the standard "length times height" area formula of a rectangle, and that is invariant under rigid motions.
Here is an outline of the proof of this theorem. One first derives the standard "one half base times height" formula for a triangle. Then one subdivides each figure into triangles and adds their areas. Finally one proves that the result is well-defined independent of subdivision; this is Lemma 23.5 in Harshorne. I've skipped some steps of the outline in which one proves special cases of well-definedness, on the way to the general proof of well-definedness.