PROPOSITION 1. THEOREM.
Triangles and parallelograms of the same altitude are to one another as their bases.
Let the triangles ABC, ACD, and the parallelograms EC, CF have the same altitude, namely, the perpendicular drawn from the point A to BD: as the base BC is to the base CD, so shall the triangle ABC be to the triangle ACD and the parallelogram EC to the parallelogram CF.
Produce BD both ways;
take any number of straight lines BG, GH, each equal to BC, and any number of straight lines DK, KL, each equal to CD; [I. 3.
and join AG, AH, AK, AL.
Then, because CB, BC, GH are all equal, [Construction.
the triangles ABC, AGB, AHG are all equal. [I. 38.
Therefore whatever multiple the base HC is of the base BC, the same multiple is the triangle AHC of the triangle ABC.
For the same reason, whatever multiple the base CL is of the base CD, the same multiple is the triangle ACL of the triangle ACD.
And if the base HC be equal to the base CL, the triangle AHC is equal to the triangle ACL; and if the base HC be greater than the base CL, the triangle AHC is greater than the triangle ACL; and if less, less. [I. 38.
Therefore, since there are four magnitudes, namely, the two bases BC, CD, and the two triangles ABC, ACD; and of the base BC, and the triangle ABC, the first and the third, any equimultiples whatever have been taken, namely, the base HC and the triangle AHC; and of the base CD and the triangle ACD, the second and the fourth, any equinmltiples whatever have been taken, namely, the base CL and the triangle ACL;
