The Elements

Four geometric diagrams are positioned in the left margin:
1. (Top) A five-pointed star (pentagram) inscribed within a pentagon with vertices labeled a, b, c, d, and e. The internal intersection points are labeled g and f. This illustrates the proof that the points of a star sum to two right angles.
2. (Upper middle) A parallelogram with vertices a, b, c, and d. A diagonal line is drawn from vertex a to vertex d. This illustrates Proposition 33 or 34.
3. (Lower middle) A parallelogram identical in structure to the one above, labeled a, b, c, and d with a diagonal line a-d.
4. (Bottom) A diagram showing two parallelograms sharing a common base line c-e. They are positioned between two parallel horizontal lines. The top line contains points labeled a, g, f, h, and b. Various lines connect these points to the base points c and e, forming overlapping figures. This illustrates Proposition 35, regarding parallelograms on the same base.

...and the others; wherefore the interior and exterior angles of the points a, b, c, d, and e are equal to ten right angles. Therefore, having subtracted the interior angles, which are equal to six right angles In Euclidean geometry, the sum of interior angles of a pentagon is always 540 degrees, or 6 right angles, the exterior ones—namely b.a.l, c.b.f, d.c.g, e.d.b, and a.e.k—will be equal to four right angles.

¶ It also appears that for every pentagon in which each side intersects two of the remaining sides This refers to a pentagram or star-pentagon, it has five angles equal to two right angles. Let the proposed pentagon be a.b.c.d.e. and let side a.c. cut side b.e. at point g, and side a.d. cut the same side b.e. at point f. Angle a.f.g. will then be equal to the two angles b. and d., because it is an exterior angle to them in triangle f.d.b. Likewise, angle f.g.a. will be equal to the two angles c. and e., since it is an exterior angle to them in triangle g.c.e. But the two angles a.f.g. and f.g.a. together with angle a. are equal to two right angles as they form the three angles of the inner triangle a.f.g.; therefore, the four angles b., d., and c., e. are, together with angle a., equal to two right angles: which is what was proposed.

Proposition 33.

If two lines of equal length and parallel to each other are joined at their endpoints by two other lines, those lines will also be equal and parallel.

¶ Let there be two lines, a.b. and c.d., which are equal and parallel, whose endpoints I join by the lines a.c. and b.d., which I say are also equal and parallel. I shall draw line a.d.; and because lines a.b. and c.d. are parallel, angle b.a.d. will be equal to angle a.d.c. by the first part of Proposition 29 regarding alternate interior angles. Therefore, the two sides a.b. and a.d. of triangle a.b.d. will be equal to the two sides d.c. and d.a. of triangle d.c.a., and angle a. of the first triangle is equal to angle d. of the second. Thus, by Proposition 4 the Side-Angle-Side theorem, the base b.d. of the first is equal to the base a.c. of the second, and angle a.d.b. of the first is equal to angle d.a.c. of the second. And since these are alternate angles, lines b.d. and a.c. will be parallel by Proposition 27; and since it was previously proven that they are equal, both parts of the proposition are clear.

Proposition 34.

Every surface enclosed by parallel sides a parallelogram has its opposite sides and angles equal to one another, and its diagonal divides it through the middle.

¶ Let there be a surface a.b.c.d. with parallel sides, such that line a.b. is parallel to c.d. and a.c. is parallel to b.d. I say that the two lines a.b. and c.d., and likewise the two lines a.c. and b.d., are equal. Similarly, I say that angle a. is equal to angle d., and angle b. to angle c. I shall draw the diagonal a.d., which will also divide that surface in the middle. Since a.b. and c.d. are parallel, the angles b.a.d. and c.d.a., which are alternate angles, will be equal by Proposition 29. And because a.c. and d.b. are also parallel, the angles c.a.d. and b.d.a., which are alternate, will be equal by the same proposition. I consider, then, the two triangles a.d.b. and d.a.c.; and because two angles a. and d. of triangle a.b.d. are equal to two angles d. and a. of triangle d.a.c., and side a.d. upon which those angles lie is common to both triangles, it follows by Proposition 26 the Angle-Side-Angle theorem that side a.b. is equal to side c.d., and side a.c. to side b.d., and angle b. to angle c. And since the total angle a. is shown to be equal to the total angle d. by the second common notion adding equals to equals results in equals, the whole proposition with its corollary is evident.

Proposition 35.

All parallelograms original: "superficies equidistantium laterum" established upon one base and within the same parallel lines are proven to be equal to each other.

¶ Let there be two parallel lines, a.b. and c.d., between which the parallelogram a.c.f.e. is formed on base c.e.; and on the same base and between the same lines, let another parallelogram g.c.b.e. be formed. I say that the two aforementioned surfaces are equal, which is proved thus: for either line c.g. will cut line a.b. at some point on line a.f., or at point f., or at some point on line b.f. First, let it cut at some point on line a.f., as appears in the first figure. And because each of the two lines a.f. and g.b. is equal to line c.e. by the preceding proposition, one of them will be equal to the other...