1. The upper side of a piece of paper lying flat upon a table is a plane surface, and so is the lower side which is in contact with the table.

2. The two surfaces are separated by the material of the paper. The material being very thin, the other sides of the paper do not present appreciably broad surfaces, and the edges of the paper are practically lines. The two surfaces though distinct are inseparable from each other.

3. Look at the irregularly shaped piece of paper shown in Fig. 3, and at this page which is rectangular. Let us try and shape the former paper like the latter.

4. Place the irregularly shaped piece of paper upon the table, and fold it flat upon itself. Let X'X be the crease thus formed. It is straight. Now pass a knife along the fold und separate the smaller piece. We thus obtain one straight edge.

5. Fold the paper again as before along BY, so that the edge X'X is doubled upon itself. Unfolding the paper, we see that the crease BY is at right angles to the edge X'X. It is evident by superposition that the angle YBX' equals the angle XBY, and that each of these angles equals an angle of the page. Now pass a knife as before along the second fold and remove the smaller piece.

I The Square 3

Fig. 3.

6. Repeat the above process and obtain the edges CD and DA. It is evident by superposition that the angles at A, B, C, D, are right angles, equal to one another, and that the sides BC, CD are respectively equal to DA, AB. This piece of paper (Fig. 3) is similar in shape to the page.

I The Square 4

Fig. 4.

7. It can be made equal in size to the page by taking a larger piece of paper and measuring off AB and BC equal to the sides of the latter.

8. A figure like this is called a rectangle. By superposition it is proved that (1) the four angles are right angles and all equal, (2) the four sides are not all equal, (3) but the two long sides are equal, and so also are the two short sides.

9. Now take a rectangular piece of paper, A'B'CD, and fold it obliquely so that one of the short sides, CD, falls upon one of the longer sides, DA', as in Fig. 4. Then fold and remove the portion A'B'BA which overlaps. Unfolding the sheet, we find that ABCD is now square, i. e., its four angles are right angles, and all its sides are equal.

I The Square 5

Fig. 5.

10. The crease which passes through a pair of the opposite corners B, D, is a diagonal of the square. One other diagonal is obtained by folding the square through the other pair of corners as in Fig. 5.

11. We see that the diagonals are at right angles to each other, and that they bisect each other.

12. The point of intersection of the diagonals is called the center of the square.

I The Square 6

Fig. 6.

13. Each diagonal divides the square into two congruent right-angled isosceles triangles, whose vertices are at opposite corners.

14. The two diagonals together divide the square into four congruent right-angled isosceles triangles, whose vertices are at the center of the square.

15. Now fold again, as in Fig. 6, laying one side of the square upon its opposite side. We get a crease which passes through the center of the square. It is at right angles to the other sides and (1) bisects them; (2) it is also parallel to the first two sides; (3) it is itself bisected at the center; (4) it divides the square into two congruent rectangles, which are, therefore, each half of it; (5) each of these rectangles is equal to one of the triangles into which either diagonal divides the square.

I The Square 7

Fig. 7.

16. Let us fold the square again, laying the remaining two sides one upon the other. The crease now obtained and the one referred to in § 15 divide the square into four congruent squares.

17. Folding again through the corners of the smaller squares which are at the centers of the sides of the larger square, we obtain a square which is inscribed in the latter. (Fig. 7.)

I The Square 8

Fig. 8.

18. This square is half the larger square, and has the same center.

19. By joining the mid-points of the sides of the inner square, we obtain a square which is one-fourth of the original square (Fig. 8). By repeating the process, we can obtain any number of squares which are to one another as

½,¼,⅛,1/16, etc., or ½,1/22,1/23,1/24,.....

Each square is half of the next larger square, i. e., the four triangles cut from each square are together equal to half of it. The sums of all these triangles increased to any number cannot exceed the original square, and they must eventually absorb the whole of it.

Therefore ½ + 1/22 + 1/23 + etc. to infinity = 1.

20. The center of the square is the center of its circumscribed and inscribed circles. The latter circle touches the sides at their mid-points, as these are nearer to the center than any other points on the sides.

21. Any crease through the center of the square divides it into two trapezoids which are congruent. A second crease through the center at right angles to the first divides the square into four congruent quadrilaterals, of which two opposite angles are right angles. The quadrilaterals are concyclic, i. e., the vertices of each lie in a circumference.