The Elements (1482) - Page 15

Three geometric diagrams are placed in the right margin, corresponding to the propositions in the text: 1\. (Prop. 26) Two triangles are shown. The first is labeled 'a' at the top, 'b' at the bottom left, and 'c' at the bottom right. The second triangle is labeled 'd' at the top, 'e' at the bottom left, and 'f' at the bottom right. A point 'g' is marked on the side 'de' of the second triangle, and a line 'gf' is drawn to illustrate a proof by contradiction. 2\. (Prop. 27) Two horizontal lines are intersected by a diagonal transversal line. The top horizontal line is labeled 'c' on the left and 'd' on the right. The bottom horizontal line is labeled 'e' on the left and 'f' on the right. The transversal is labeled 'k' at the top and 'b' at the bottom. The intersection points are labeled 'g' (on line cd) and 'b' (on line ef). 3\. (Prop. 28) Similar to the previous diagram, two horizontal lines are intersected by a transversal. The horizontal lines are labeled 'c-d' (top) and 'e-f' (bottom). The transversal is labeled 'a' at the top and 'b' at the bottom. The intersection points are labeled 'g' (top) and 'b' (bottom).

For any two triangles in which two angles of the one are equal to two angles of the other, and one side of the one is equal to one side of the other—whether that side is between the equal angles or opposite one of them—the remaining two sides of the first triangle will be equal to the remaining two sides of the second triangle, each to its corresponding side, and the remaining angle of the first will be equal to the remaining angle of the second.

¶ Let there be two triangles, .a.b.c. and .d.e.f. Let angle .b. be equal to angle .e. and angle .c. be equal to angle .f. Also, let side .b.c. be equal to side .e.f., or let one of the other two sides, .a.b. and .a.c., be equal to one of the other two sides, .d.e. and .d.f., such that .a.b. is equal to .d.e. or .a.c. is equal to .d.f. I say that the remaining two sides of the one will be equal to the remaining two sides of the other, and the remaining angle will be equal to the remaining angle (angle .a. equal to angle .d.).

First, let the side .b.c. (on which angles .b. and .c. lie) be equal to side .e.f. (on which angles .e. and .f. lie). Then I say that side .a.b. is equal to side .d.e., side .a.c. is equal to side .d.f., and angle .a. is equal to angle .d. For if side .a.b. is not equal to side .d.e., one of them will be greater. Let .d.e. be the greater, which I shall cut to be equal to .a.b. Let .g.e. be equal to .a.b., and I will draw the line .g.f. By Proposition 4 Proposition 4 is the Side-Angle-Side theorem, which states that if two sides and the included angle are equal, the triangles are congruent., angle .g.f.e. would then be equal to angle .a.c.b. Since angle .a.c.b. was said to be equal to angle .d.f.e., this would mean the part is equal to the whole original: "ps toti" (pars toti). Euclid treats this as a logical impossibility or "common notion" that the part cannot equal the whole., which is impossible. Therefore, .d.e. must be equal to .a.b. Consequently, by Proposition 4, .d.f. is equal to .a.c. and angle .d. is equal to angle .a. This concludes the first part of the division.

Again, as before, let the two angles .b. and .c. be equal to the two angles .e. and .f., and let side .a.b. (which is opposite angle .c.) be equal to side .d.e. (which is opposite angle .f., to which angle .c. was set as equal). I say that side .b.c. will be equal to side .e.f., side .a.c. to side .d.f., and angle .a. to angle .d. For if side .e.f. were not equal to side .b.c., one would be greater. Let .e.f. be the greater. Let .e.g. be set equal to .b.c., and I will draw line .d.g. By Proposition 4, angle .d.g.e. will be equal to angle .a.c.b., and therefore to angle .d.f.e. This would mean an exterior angle An angle formed outside a triangle by extending one of its sides. is equal to an interior angle An angle inside the triangle., which is impossible by Proposition 16 Proposition 16 states that an exterior angle of a triangle is always greater than either of the opposite interior angles.. Therefore, .e.f. must be equal to .b.c. Consequently, by Proposition 4, side .d.f. is equal to side .a.c., and the total angle .d. is equal to angle .a. This is the second part of the division; thus the whole is clearly proven.

Proposition 27.

If a straight line falls upon two straight lines and makes the alternate angles Angles on opposite sides of the transversal line, but inside the two lines being intersected. equal to one another, those two lines will be parallel original: "equidistantes". Lines that never meet, no matter how far they are extended..

¶ Let it be that line .a.b. falls upon two lines .c.d. and .e.f., and cuts line .c.d. at point .g. and line .e.f. at point .b. In the diagram, this point is labeled 'b', though the transversal is also called 'ab'. Let angle .d.g.b. be equal to angle .e.b.g. I say that lines .c.d. and .e.f. are parallel. For if they were not, they would meet original: "pcurrant" (procurrant). either on the side of .c. and .e. at point .k., or on the side of .d. and .f. at point .l. Whichever the case, an impossibility would occur by Proposition 16: namely, that an exterior angle would be equal to an interior angle. For one of the said alternate angles, which were set as equal, would be an exterior angle and the other an interior angle of the resulting triangle. Since it is therefore impossible for them to meet when extended in either direction, they will be parallel by definition; which is what was proposed.

Proposition 28.

If a straight line comes upon two straight lines and its exterior angle is equal to the interior and opposite angle on the same side, or if the two interior angles on the same side are equal to two right angles, those two lines will be parallel.

¶ Let it be that line .a.b. cuts two lines .c.d. and .e.f. at points .g. and .b. Let the exterior angle .g. be equal to the interior angle .b. taken on the same side; or let the two interior angles .g. and .b. taken on the same side be equal to two right angles...