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 $i \le n$?