The Elements (1482) - Page 26

And by Proposition 2 of this book original: p.2.huius, it follows that the area formed by the whole line a.b. multiplied by itself is equal to the area formed by the same line multiplied by segments a.c. and c.b. But by the whole line multiplied by a.c., there is produced exactly as much as from a.c. multiplied by itself and from a.c. multiplied by b.c., according to Proposition 3 of this book. Likewise, from the whole line a.b. multiplied by b.c., there is produced exactly as much as from c.b. multiplied by itself and from c.b. multiplied by a.c., by the same proposition. Therefore, the area formed from the whole a.b. multiplied by itself is equal to the area formed by a.c. multiplied by itself and by c.b., and by c.b. multiplied by itself and by a.c., which is what was proposed. However, by this path the conclusion is not as clear as it is by the preceding path; wherefore the first method is more in harmony with the author's intent.

Proposition 5

A geometric diagram showing a large rectangle subdivided by vertical lines and a horizontal line. A diagonal line runs from the top left of the inner rectangle (e) to the bottom right (b). Points are labeled a, c, d, b on the bottom edge; k, l, h, m across the middle; and e, g, f across the top. This represents the "Gnomon" proof for Proposition 5.

If a straight line be cut into two equal and two unequal parts, the rectangleoriginal: rectangulum — a four-sided figure with right angles contained by the unequal segments of the whole, together with the square described on the segment between the two points of section, is equal to the square described on half of the whole line.
¶ Let line a.b. be divided equally at point c. and unequally at point d. I say that the square of c.b. is equal to the rectangle formed by a.d. multiplied by d.b. and the square of c.d. ¶ I shall describe the square of c.b., which is c.b.f.e., in which I shall draw the diameter e.b. and draw d.g. parallel to b.f., which intersects the diameter e.b. at point h. And from point h. I shall draw a line parallel to line a.b., which is h.k., intersecting line b.f. at point m. and line c.e. at point l. And I shall draw a.k. parallel to c.e. By the corollary of the preceding [proposition], both of the two surfaces l.g. and d.m. will be squares. And by Proposition 43 of the First Book, the two complementsoriginal: supplementa — the areas remaining in a parallelogram when squares are drawn along the diameter c.h. and h.f. are equal. Therefore, by adding the square d.m. to each, the parallelogram c.m. will be equal to the parallelogram d.f. And because a.l. is equal to c.m. by Proposition 36 of the First Book, the rectangle a.b. The text here refers to the area created by the unequal segments will be equal to the gnomonoriginal: gnomonem — an L-shaped figure remaining after a smaller square is removed from a larger one which surrounds the square l.g. Therefore, by adding the square l.g. to each, the rectangle a.b. with the square l.g. will be equal to the square e.f., which is what was proposed.

Proposition 6

A geometric diagram showing a line a-b extended to point d. A large rectangle is constructed above it, divided into sections by vertical and horizontal lines, with a diagonal running from point e to point d. Points are labeled a, c, b, d along the bottom; k, l, h, m across the middle; and e, g, f along the top.

If a straight line be bisected, and another straight line be added to it in a straight line, the rectangle contained by the whole line made up of the original and the added part, multiplied by the added part, together with the square on the half, is equal to the square described on the line made up of the half and the added part.
¶ Let line a.b. be divided equally at point c., and let line b.d. be added to it. I say that the square of c.d., which is c.d.e.f., is equal to the rectangle formed by the whole a.d. multiplied by b.d. and the square of c.b. I shall draw the diameter d.e. in the aforesaid square and draw line b.g. parallel to d.f., which intersects the diameter d.e. at point h. From this point h. I shall draw a line parallel to line a.b., which is b.k. Likely h.k. in the diagram, intersecting d.f. at point m. and c.e. at point l. And I shall draw a.k. parallel to c.l. By Proposition 36 of the First Book, a.l. will be equal to c.b. referring to the area. But c.b. will be equal to h.f. by Proposition 43 of the First Book. Wherefore a.l. is equal to h.f. Therefore, by adding c.m. to both sides, a.m. will be equal to the whole gnomon surrounding l.g. Wherefore, by adding l.g. to both sides, a.m. with l.g. will be equal to the whole square c.f. And because both of the two surfaces l.g. and b.m. are squares, the proposition is clear by the corollary of Proposition 4 of this book.

Proposition 7

A geometric diagram of a square with a diagonal line from top left to bottom right. The square is subdivided into four sections. Points are labeled a, c, b along the bottom; h, g, f across the middle; and e, d at the top.

If a line be divided into two parts, the square of the whole line multiplied by itself, together with the square of one of the parts, is equal to twice the rectangle contained by the whole line and that same part, together with the square of the other part.
¶ Let line a.b. be divided into two parts at point c. I say that the square of the whole a.b. with the square of b.c. is equal to twice the rectangle formed by a.b. and b.c., together with the square of a.c. Let the square of the whole be described as a.b.d.e. and let the diameter b.d. be drawn and...

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