The Elements

...along line a.d. I construct angle e.a.d. by the instruction of Proposition 23 original: "p doctrinā.23.", equal to angle b.d.a., its alternate; and line a.e. will be parallel to b.c. by Proposition 27, which is what was proposed.

Proposition 32.

In any triangle, an exterior angle is equal to the sum of the two interior and opposite angles; and it is necessary that all three interior angles of the triangle are equal to two right angles.

¶ Let there be a triangle a.b.c. whose side b.c. is extended to d. I say that the exterior angle at c. the angle a.c.d. is equal to the two interior and opposite angles a. and b. joined together; and that the three angles of triangle a.b.c. joined together are equal to two right angles. From point c., I shall draw line c.f. parallel to a.b. following the instruction of the preceding proposition. Angle f.c.a. will be equal to angle a. because they are alternate angles original: "coalterni" by the first part of Proposition 29; and the exterior angle f.c.d. is equal to the interior angle b. by the second part of the same proposition. Therefore, the whole exterior angle a.c.d. is equal to the two interior and opposite angles a. and b.; which is the first part of the proof. And since the two angles a.c.b. and a.c.d. are equal to two right angles by Proposition 13 which states that angles on a straight line sum to 180 degrees, it follows that the three interior angles a., b., and c. will be equal to two right angles; which is the second thing proposed.

¶ From this, moreover, it is clear that for any polygonal figure original: "figure poligonie", the sum of all its angles is equal to twice as many right angles as its position in the sequence of polygons, starting from the first. For example, among polygonal figures, the triangle is the first; for if a figure were made of only two lines—since a figure is an enclosure of lines—then two straight lines would enclose a surface, which is impossible by the last postulate. The quadrilateral is the second figure; the pentagon is the third; and similarly, any figure's place in the order will be the number of its sides or angles minus two original: "dempto binario".

I say, therefore, that for the triangle, which is first, all angles are equal to two right angles. For the quadrilateral, which is second, they will be equal to four right angles; and for the pentagon, which is third, they will be equal to six right angles. This is manifest from the fact that any such figure can be resolved into as many triangles as its position from the first, by drawing straight lines from any of its angles to all the opposite angles. Since all the angles of every triangle are equal to two right angles, all the angles of any many-sided figure will be equal to twice as many right angles as its position from the first; which is what was proposed.

For example, let there be a pentagon a.b.c.d.e. From its angle a., I shall draw lines to the opposite angles c. and d. The whole pentagon will be resolved into three triangles: a.b.c., a.c.d., and a.d.e. Since the angles of each of these are equal to two right angles, the angles of the pentagon will be equal to six right angles; which is double the number of its distance from the first 3rd position x 2 = 6, or double the number of its angles or sides minus two (5 sides - 2) x 2 = 6.

¶ We can also propose the same thing this way, saying that the sum of all angles of any polygonal figure is equal to twice as many right angles as it has angles, minus four. If any point is marked within the figure and lines are drawn from it to each angle, the figure itself will be resolved into as many triangles as it has angles. Therefore, all the angles of all those triangles taken together will be equal to twice as many right angles as there are angles in the figure. Since all the angles of the triangles meeting around the central point are equal to four right angles by Proposition 13 as they complete a full circle, the proposition is clearly manifest.

¶ Similarly, it is also clear that the sum of all the exterior angles of any polygonal figure is equal to four right angles. For the interior and exterior angles together are equal to twice as many right angles as there are angles [of the figure], by Proposition 13. But the interior angles are equal to twice as many right angles as there are angles minus four; therefore, the exterior angles must be equal to four right angles; which is what was proposed.

For example, let the sides of the proposed pentagon be extended so that exterior angles are formed: side a.b. extended to f., b.c. to g., c.d. to h., d.e. to k., and e.a. to l. By Proposition 13, the two angles—interior a. and exterior a.—will be equal to two right angles. By the same reasoning, the two angles—interior b. and exterior b.—and so on.