The Elements

In every triangle, the longer side is opposite the greater angle.
¶ Let it be as in triangle .a.b.c., where side .b.c. is greater than side .a.b. I say that angle .a. will be greater than angle .c. This is the converse of the preceding proposition The "preceding proposition" refers to Proposition 18.. For if it were equal, then by Proposition 6, side .a.b. would be equal to side .b.c., which is contrary to the hypothesis. If, however, angle .c. were greater, then by the preceding [proposition], side .a.b. would be greater than side .b.c., which is also contrary to the hypothesis; wherefore the proposition is proved.

In every triangle, any two sides joined together are longer than the remaining one.
¶ Let there be triangle .a.b.c. I say that the two sides .a.b. and .a.c. are longer than side .b.c. Extend line .b.a. as far as .d. so that .a.d. is equal to .a.c., and draw line .c.d. By Proposition 5, angle .a.c.d. will be equal to angle .d.; wherefore angle .b.c.d. is greater than angle .d. Therefore, by Proposition 18 original: "p.18."; Note: The text cites Proposition 18, though in modern numbering this logic follows Proposition 19, stating that the longer side is opposite the larger angle., side .b.d. is greater than side .b.c. But .b.d. is equal to .a.b. and .a.c. [combined]; wherefore .b.a. and .a.c. joined together are greater than .b.c.

If from the two endpoints of one side of a triangle, two lines drawn inside the triangle itself meet at a single point, these same lines will be shorter than the other two sides of the triangle, but will contain a greater angle.
¶ Let it be that in triangle .a.b.c., from the extremities of side .b.c., two lines .b.d. and .c.d. meet at point .d. inside triangle .a.b.c. I say that joined together, they are shorter than the two lines .a.b. and .a.c. joined together, and that angle .d. is greater than angle .a. For I shall extend .b.d. until it cuts side .a.c. at point .e. By Proposition 20, .b.a. and .a.e. joined together are greater than .b.e. Therefore, .b.a. and .a.c. Since side .a.c. is composed of .a.e. and .e.c. are greater than .b.e. and .e.c. But indeed, .d.e. and .e.c. joined together are greater than .d.c. by the same [proposition]; wherefore .b.e. and .e.c. are greater than .b.d. and .d.c. Because .b.e. is the sum of .b.d. and .d.e.. And because .b.a. and .a.c. are greater than .b.e. and .e.c. as was proved before, they will be, much more certainly, greater than .b.d. and .d.c., which is the first part of the proposition. Now, since angle .b.d.c. is greater than angle .d.e.c. by Proposition 16, and angle .d.e.c. is greater than angle .c.a.b. by the same; angle .b.d.c. will be much more certainly greater than angle .b.a.c., which is the second part of the proposition.

Given three straight lines, any two of which joined together are longer than the remaining one, to construct a triangle from three other lines equal to them.
¶ Let the three proposed straight lines be .a. .b. .c., and let any two joined together be longer than the remaining one. For otherwise, a triangle could not be constructed from three lines equal to them by Proposition 20. Since, therefore, I wish to construct a triangle from those three aforementioned lines: I take a straight line which is .d.e., to which I do not set a determined end on the side of .e. From this, by Proposition 3, I take .d.f. equal to .a., and .f.g. equal to .b., and .g.h. equal to .c. Secondly, with point .f. as center, I describe a circle .d.k. according to the length of line .f.d. Likewise, having made .g. the center, I describe [circle] .k.h. according to the length of line .g.h. These circles will intersect each other at two points, one of which is .k. Otherwise, it would follow that one of said lines is equal to the other two joined together or greater than them, which is contrary to what was posited. I therefore draw lines .k.f. and .k.g. Thus, triangle .k.f.g. is constructed from three lines equal to the given lines .a. .b. .c. For .f.d. and .f.k. are equal because they [radiate] from the center to the circumference; wherefore .f.k. is equal to .a. Similarly, .g.h. and .g.k. are equal because they go from the center to the circumference; wherefore .g.k. is equal to .c. And because .g.f. was taken equal to .b., the proposition is clearly manifest.

Vocabulary: Euclid, geometry, triangle, proposition, circle, angle, side