Triangles and parallelograms which are of the same height are to one another as their bases. * Let $ABC$ and $ACD$ be triangles, and $EC$ and $CF$ parallelograms, of the same height $AC$. * I say that as base $BC$ is to base $CD$, so triangle $ABC$ (is) to triangle $ACD$, and parallelogram $EC$ to parallelogram $CF$.
The proportion of areas of triangles (parallelograms) of the same height is equal to the proportion of the lengths of their bases.
Proofs: 1
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As is easily demonstrated, this proposition holds even when the triangles, or parallelograms, do not share a common side, and/or are not right-angled (translator's note). ↩