The Elements

...triangles which are $b, k$ and $c, f$. original: "triangulū que sunt.b.k.z.c.f.z" — The 'z' characters likely represent 'et' (and) or punctuation in the transcription. Since each of the two angles $bac$ and $bag$ is a right angle, then by Proposition 14Proposition 14 states that if two lines make adjacent angles equal to two right angles, they form a single straight line., $gc$ will be a single straight line. For the same reason, $bh$ will be a single line, because each of the two angles $cab$ and $cah$ is a right angle. Therefore, because the parallelogram $bfga$ and the triangle $bfc$ are established upon the base $bf$ and between the two parallel lines original: "equidistantes" which are $cg$ and $bf$, then by Proposition 41, the parallelogram $bfga$ will be double the triangle $bfc$. Furthermore, triangle $bfc$ is equal to triangle $abd$ by Proposition 4The 'Side-Angle-Side' theorem of congruence., because the sides $fb$ and $bc$ of the first are equal to the sides $ab$ and $bd$ of the latter, and the angle $b$ of the first is equal to angle $b$ of the latter, for each consists of a right angle and the common angle $abc$. Therefore, the parallelogram $bfga$ is double the triangle $abd$. But the parallelogram $bdlm$ is also double the same triangle by Proposition 41, because they are established upon the same base, namely $bd$, and between the parallel lines $bd$ and $al$. Therefore, by common scienceoriginal: "cōmunē sciām" — referring to Euclid’s Common Notions or Axioms; specifically, things equal to the same thing are equal to each other., the square $abfg$ and the parallelogram $bdlm$ are equal, because their halves (namely the aforementioned triangle) are equal. In the same manner and through the same propositions, by means of the triangles $kbc$ and $aec$, we shall prove that the square $ackb$ is equal to the parallelogram $celm$. Thus the proposition is clear.

Proposition 47

A geometric diagram in the right margin illustrates the converse of the Pythagorean Theorem (Proposition 47). It depicts a triangle with a perpendicular line dropped from the apex to the base. The vertices are labeled 'c' at the top, 'a' at the bottom left, 'b' at the intersection of the perpendicular on the base, and 'd' at the bottom right.

If the product of one side of a triangle multiplied by itself A literal translation of the Latin phrase for squaring a line. is equal to the two squares which are described from the two remaining sides, then the angle which that side opposes is a right angle.
¶ To lead a line into itself is to describe its square. ¶ Let there be a triangle $abc$, and let the square of side $ac$ be equal to the squares of the two sides $ab$ and $bc$ joined together. I say that angle $b$, which side $ac$ opposes, is a right angle; and this is the converse of the previous theorem. ¶ From point $b$, I draw the line $bd$ perpendicular to line $bc$, which I place equal to $ab$, and I extend the line $dc$. By the preceding theorem, the square of $dc$ will be equal to the two squares of the two lines $db$ and $bc$. And because $bd$ was placed equal to $ba$, it follows by common science—which is that the squares of equal lines are equal—that the squares of the two lines $ab$ and $bd$ are equal. For this reason, the square of $dc$ will be equal to the square of $ac$. Therefore, by another common science—which is the converse of the first, namely that lines whose squares are equal are themselves equal—$dc$ will be equal to $ac$. Wherefore, by Proposition 8, angle $b$ of triangle $abc$ is a right angle, which is what was to be proposed.

Proposition 48

A geometric diagram in the right margin illustrating Proposition 48, showing the construction of a gnomon. It includes two nested squares; the inner square is labeled 'a'. The surrounding gnomon and extended construction are labeled with points 'b' (bottom left), 'f', 'c' (bottom right), 'd', and 'e'.

Given any squares, to describe around one of them a gnomonA gnomon is the L-shaped figure remaining after a smaller square is removed from a larger one. equal to the other square.
¶ Let two squares be proposed, namely $ab$ and $cd$, and let the proposal be to produce a gnomon around $ab$ equal to the square $cd$. Therefore, let one side of square $ab$ be extended to the length of one side of square $cd$ in a continuous and straight line, and let this be $fe$, so that $fe$ is equal to one of the sides of square $cd$. And from $e$, I shall lead a straight line to $a$. Let there be, therefore, a right-angled original: "orthogoni⁹" triangle because $f$ is a right angle. Let it be argued, therefore, according to the penultimate [proposition] of the first book Referring back to the Pythagorean Theorem. thus: the square $ea$ is as much as square $ef$ and square $fa$. But square $ef$ is equal to square $cd$, and square $fa$ is equal to square $ab$. Therefore, square $ae$ is equal to squares $ab$ and $cd$. Likewise, $efa$ is a triangle, therefore sides $ef$ and $fa$ are longer than side $ae$, according to Proposition 20 of the first book. But $fa$ is equal to $ab$ by reason of the square; therefore $ef$ and $fb$ are longer than $ae$. Therefore that total line, namely $eb$, is greater than $ae$. Let $be$ therefore be cut back to the equality of $ae$ at point $c$, so that $bc$ is equal to $ae$. Therefore the square $bc$ is equal to the square $ae$. But square $ae$, as was previously proved, is equal to squares $ab$ and $cd$. Therefore square $bc$ is equal to the same. But square $bc$ adds upon square $ab$ that gnomon which you see. Therefore the gnomon...

A hand-drawn marginal sketch in the bottom-left corner. It shows a triangle with vertices labeled A, B, and C, with auxiliary construction lines extending to points labeled D, E, F, and G. A large 'X' mark is located below the sketch.