The Elements

Four geometric diagrams are positioned in the left margin, corresponding to the text:
1. (Top) A triangle labeled a-b-c. A line d-e is drawn parallel to the base b-c, intersecting sides a-b and a-c. A marginal note "Corollary" is written next to vertex b.
2. (Upper middle) A complex diagram showing two triangles on equal bases (b-c and e-f) along the same line. Parallel lines are drawn through the bases and the vertices. Points g, a, and h are on the top line; d is a vertex; b, c, e, and f are on the bottom line.
3. (Lower middle) A parallelogram a-b-c-d. A triangle e-b-d is drawn sharing the same base b-d and between the same parallel lines. A marginal note "Corollary" is written to the right of point d.
4. (Bottom) A parallelogram with various internal lines including diagonals and constructed triangles, illustrating the relationship between parallelograms and triangles of the same base and height.

...line a-e at point e, and I will draw line e-c. And because triangle e-b-c is equal to triangle a-b-c by Proposition 37, and triangle d-b-c was stated to be equal to triangle a-b-c, then triangle d-b-c will be equal to triangle e-b-c—the part being equal to the whole original: "pars toti," a logical fallacy where a portion is claimed to be identical to the entire entity, which is impossible. Therefore, the line which is drawn parallel to b-c from point a will not pass above d. Let it then pass below, and let it be a-f, cutting line d-b at point f. I will then draw line f-c.

Corollary

And because by Proposition 37 triangle f-b-c is equal to triangle a-b-c, it will also be equal to triangle d-b-c—the part being equal to the whole, which is impossible. Because, therefore, the line from point a parallel to b-c does not pass through any point except d, the proposition is clear. From this and the preceding, note that if any straight line bisects original: "per equalia secet," to cut into two equal parts two sides of any triangle, that line will be parallel to the base. This is proved as follows: Let there be a triangle a-b-c whose two sides a-b and b-c are bisected by line d-e—a-b at point d and b-c at point e. I say that line d-e is parallel original: "equidistans" to a-c. For I will draw the diagonals a-e and d-c in the quadrilateral a-c-e-d. By Proposition 38, triangle a-e-d will be equal to triangle d-e-b, because line a-d was established as equal to line d-b. Likewise, by the same proposition, triangle c-e-d will be equal to the same triangle d-e-b, because line c-e was established as equal to line e-b. Since triangle a-e-d is equal to triangle c-e-d, and because they are established upon the same base—namely, line e-d—and on the same side, they will be between parallel lines by this Proposition 39. Therefore, line d-e is parallel to line a-c, which proposition will be useful to you for the fifth proposition of the fourth book.

Let there be two equal triangles, a-b-c and d-e-f, established upon two bases which are b-c and e-f, and on the same side. I say they are between two parallel lines. This is the converse of Proposition 38, and it is proved by it just as the preceding was proved by Proposition 37. From point a, let a line be drawn parallel to line b-f. If it passes through point d, the proposition is clear. If, however, it passes above it, as line a-g, and e-d is extended to it at point g, and line g-f is drawn, then by Proposition 38 triangle a-b-c will be equal to triangle g-e-f. Wherefore triangle d-e-f will be equal to triangle g-e-f—the part being equal to the whole, which is impossible. Therefore, it shall not pass above. Let it then pass below, and let it cut line d-e at point h, and let line f-h be drawn. Then by Proposition 38 triangle b-e-f The text likely means a-b-c here will be equal to triangle a-b-c, and therefore also to triangle d-e-f—the part being equal to the whole, which is impossible. Because, therefore, it will not pass except through point d, the proposition is clear.

Let there be a parallelogram a-b-c-d and triangle e-b-d on the base b-d and between
Corollary

lines a-e and b-d, which are parallel. I say the parallelogram is double the triangle. Draw the diagonal a-d in the parallelogram. Triangle a-b-d will be half of the parallelogram by the conclusion original: "conclusione" of Proposition 34. And because triangle e-b-d is equal to triangle a-b-c likely error for a-b-d by Proposition 37, it is clear that triangle e-b-d is half of the parallelogram a-b-c-d, which is what was proposed. Similarly, it can also be proved that if a parallelogram and a triangle are established on equal bases and between parallel lines, the parallelogram will be double the triangle. Euclid did not set this down separately because it is easily seen as a corollary A proposition that follows naturally from one already proved. from this and Proposition 38, by dividing the parallelogram by a diagonal into two triangles, either upon the base of the parallelogram or between the same parallel lines...