#### PROPOSITION 1.

Triangles and parallelograms which are under the same height are to one another as their bases.

Let ABC, ACD be triangles and EC, CF parallelograms under the same height;
I say that, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD, and the parallelogram EC to the parallelogram CF.

For let BD be produced in both directions to the points H, L and let [any number of straight lines] BG, GH be
made equal to the base BC, and any number of straight lines DK, KL equal to the base CD; let AG, AH, AK, AL be joined.

Then, since CB, BG, GH are equal to one another,

the triangles ABC, AGB, AHG are also equal to one another. [I. 38]

Therefore, whatever multiple the base HC is of the base BC, that multiple also is the triangle AHC of the triangle ABC.

For the same reason,
whatever multiple the base LC is of the base CD, that multiple also is the triangle ALC of the triangle ACD; and, if the base HC is equal to the base CL, the triangle AHC is also equal to the triangle ACL, [I. 38] if the base HC is in excess of the base CL, the triangle AHC
is also in excess of the triangle ACL, and, if less, less.

Thus, there being four magnitudes, two bases BC, CD and two triangles ABC, ACD, equimultiples have been taken of the base BC and the
triangle ABC, namely the base HC and the triangle AHC, and of the base CD and the triangle ADC other, chance, equimultiples, namely the base LC and the triangle ALC;

and it has been proved that,
if the base HC is in excess of the base CL, the triangle AHC
is also in excess of the triangle ALC; if equal, equal; and, if less, less.

Therefore, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD. [V. Def. 5]

Next, since the parallelogram EC is double of the triangle
ABC, [I. 41] and the parallelogram FC is double of the triangle ACD, while parts have the same ratio as the same multiples of them, [V. 15] therefore, as the triangle ABC is to the triangle ACD, so is
the parallelogram EC to the parallelogram FC.

Since, then, it was proved that, as the base BC is to CD, so is the triangle ABC to the triangle ACD, and, as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF,
therefore also, as the base BC is to the base CD, so is the parallelogram EC to the parallelogram FC. [V. 11]

Therefore etc. Q. E. D. 1 2 3

1 Under the same height. The Greek text has “under the same height AC,” with a figure in which the side AC common to the two triangles is perpendicular to the base and is therefore itself the “height.” But, even if the two triangles are placed contiguously so as to have a common side AC, it is quite gratuitous to require it to be perpendicular to the base. Theon, on this occasion making an improvement, altered to “which are (ὅντα) under the same height, (namely) the perpendicular drawn from A to BD.” I have ventured to alter so far as to omit “AC” and to draw the figure in the usual way.

2 ABC, AGB, AHG. Euclid, indifferent to exact order, writes “AHG, AGB, ABC.”

3 Since then it was proved that, as the base BC is to CD, so is the triangle ABC to the triangle ACD. Here again words have to be supplied in translating the extremely terse Greek , literally “since was proved, as the base BC to CD, so the triangle ABC to the triangle ACD.” Cf. note on V. 16, p. 165.