The Elements

Five geometric diagrams are arranged vertically in the left margin, corresponding to the propositions in the text:
1. Two intersecting straight lines forming an "X" shape. The endpoints are labeled a, c, b, and d, with the intersection point labeled e.
2. A complex geometric construction involving a triangle (a, b, c). The side ac is bisected at point e. A line from b passes through e to a point f. Another construction shows a triangle with vertices a, b, c and an interior point g, with lines radiating to various labeled points (h, k).
3. A triangle with vertices a, b, c. The base line bc is extended to a point d.
4. A triangle with vertices a, b, c. The base line bc is extended to a point d, with a perpendicular-looking line from a to the base.
5. A triangle with vertices a, b, c. A point d is marked on the side bc, and a line is drawn from a to d, creating two smaller triangles within the larger one.

original: "ps toti: quod ē ipossibile" the part [would be equal] to the whole: which is impossible This is a fundamental "Common Notion" in Euclidean geometry: the whole is always greater than the part.; similarly, with line .c.b. extended, you will prove angle .d.b.a. to be equal to angle .f.b.a. if perhaps an adversary should say that the extended line .c.b. falls within .b.d.

For all two lines intersecting each other: all angles positioned opposite to one each other are equal; from which it is manifest that when two straight lines intersect each other, the four angles which are made are equal to four right angles.
¶ Let there be two lines .a.b. and .c.d. intersecting each other at point .e. I say that angle .d.e.b. is equal to angle .a.e.c., and angle .b.e.c. is equal to angle .a.e.d. For by the 13th [proposition], the two angles .a.e.c. and .c.e.b. will be equal to two right angles; likewise, the two angles .c.e.b. and .d.e.b. are equal to two right angles by the same [proposition]; wherefore the first two are equal to the latter two because all right angles are equal to one another by the 4th petition The 4th "petition" or Postulate: all right angles are equal.; therefore, once the common angle .c.e.b. is taken away, the angle .a.e.c. will be equal to angle .d.e.b. In the same way, it will be proven that angle .c.e.b. is equal to angle .a.e.d., which is what was proposed.

If any of the sides of a triangle be extended directly, it will make an exterior angle greater than either of the interior angles of the triangle opposite to it.
¶ Let it be that side .a.b. of triangle .a.b.c. is extended as far as .d.: I say that angle .d.b.c. is greater than each of the two interior angles opposite to it, which are .b.a.c. and .b.c.a. For I shall divide, by the 10th [proposition], line .c.b. into equal parts at point .e., and I shall extend .a.e. as far as .f., so that .e.f. is made equal to .a.e., and I shall draw line .f.b. I consider the two triangles .c.e.a. and .b.e.f.; and because the two sides .a.c. The text likely meant .a.e. and .e.c. and .e.c. of triangle .a.e.c. are equal to the two sides .f.e. and .e.b. of triangle .f.e.b., and angle .e. of the one is equal to angle .e. of the other by the preceding [proposition] because they are opposite angles; by the 4th [proposition], angle .e.c.a. will be equal to angle .e.b.f., and therefore angle .e.b.d. will be greater than angle .b.c.a. Similarly, it will also be proven that it is greater than angle .c.a.b. For let us divide .a.b. into equal parts at point .g. by the 10th [proposition], and I shall extend line g.b. Likely .g.h. in the diagram logic equal to line .c.g. by the 3rd [proposition]; afterwards, I shall draw .b.k., and there will be two triangles, which are .a.g.c. and .b.g.h. OCR reads .b.g.b.; the two sides .a.g. and .g.c. of the first are equal to the two sides .b.g. and .g.h. of the second; and angle .g. of the one [is equal] to angle .g. of the other by g.a.c. the 15th [proposition]; therefore, by the 4th [proposition], angle .g.c.a. is equal to angle .g.b.h., and therefore by the 15th, also to angle .k.b.d.; and because angle .c.b.d. is greater than angle .k.b.d., it will also be greater than angle .b.a.c., which is what was proposed.

In every triangle, any two angles are [together] less than two right angles.
¶ Let there be triangle .a.b.c. I say that any two of its angles are less than two right angles; for let one of its sides, such as .b.c., be extended as far as .d., and by the preceding [proposition], exterior angle .c. OCR reads .e. will be greater than angle .a. and greater than angle .b. But exterior [angle] .c. with interior [angle] .c. is equal to two right angles by the 13th [proposition]; therefore, interior angles .b. and .c., or interior angles .a. and .c., are less than two right angles; similarly, if side .b.a. be extended, it will be proven that the two angles .a. and .b. are less than two right angles: which is what was proposed.

In every triangle, the longer side is opposite the greater angle.
¶ Let it be that in triangle .a.b.c., angle .a. is greater than angle .c. I say that side .c.b. will be greater than side .a.b. For if they were equal, by the 5th [proposition], angle .a. would be equal to angle .c., which is [contrary] to the hypothesis; but if .a.b. were greater, let it be cut down to equality with .c.b. by the 3rd [proposition], and let .d.b. be equal to .c.b. Therefore, by the 5th [proposition], angle .d.c.b. will be equal to angle .b.d.c.; but .b.d.c. is greater than angle .b.a.c. by the 16th [proposition]; therefore .b.c.d. is greater than .b.a.c.; wherefore, much more strongly, is .a.c.b. [greater], original: "ps toto: quod ē ipossibile" the part [equal] to the whole: which is impossible.