The Elements

...of the line $ab$ in a straight line: which I place equal to the line $cf$, the base of the given triangle, upon which I constitute a single triangle equal and equilateral to it. I do this in the following manner: I establish angle $agk$ equal to angle $e$, and angle $gak$ equal to angle $f$ through Proposition 23Proposition 23 of Book I describes how to construct an angle equal to a given angle on a specific line.; and because $ga$ had been placed equal to $ef$, triangle $gak$ will be equal and equilateral to triangle $cfd$ through Proposition 26. I shall therefore divide $ga$ into equal parts at point $b$, and I shall draw line $kb$, and I shall lead from point $k$ the line $mkn$ parallel to line $gb$. Through Proposition 38, triangle $abk$ will be equal to triangle $gbk$. Then, upon point $a$ of line $ga$, I shall make angle $gal$ through Proposition 23 equal to the given angle $c$; and I shall complete, upon the base $ab$ and between the parallel lines $gb$ and $mn$, the parallelogramoriginal: "superficiem equidistantium laterum", literally "a surface of equidistant sides". $mlba$, which through Proposition 41 will be double to the triangle $kba$, and therefore equal to the total triangle $kga$, and consequently equal to the proposed triangle $def$. I shall then draw $bn$ parallel to $al$, and I shall lead the diameter $na$, which I shall draw out until it meets with $mb$ at point $o$. I shall complete the parallelogram $monq$, and I shall draw $la$ out as far as $p$. Through the preceding supplementoriginal: "supplementum". In geometry, this refers to the areas created within a parallelogram when divided by lines parallel to its sides passing through a point on the diagonal., $abpq$ will be equal to the supplement $mlba$, and therefore equal to triangle $def$. And because through Proposition 15 angle $lab$ is equal to angle $bap$, and therefore angle $bap$ is equal to angle $c$, it is clear that upon the given line $ab$ there has been described the parallelogram $abpq$ equal to the given triangle $def$, of which each of the two angles positioned opposite one another, which are $a$ and $q$, is equal to the given angle $c$, which was what was proposed.

¶ Proposition 45.

From a given line, to describe a square.
Let $ab$ be the given line from which I wish to fabricate? a square. From points $a$ and $b$ of line $ab$, I lead through Proposition 11 lines $ac$ and $bd$ perpendicular to line $ab$, which will be parallel through the last part of Proposition 28. I place each of them equal to the same $ab$ through the second [proposition], and I draw line $cd$. This line will be equal and parallel to line $ab$ through Proposition 33. And because each of the two angles $a$ and $b$ is a right angle, each of the two angles $c$ and $d$ will be a right angle through the last part of Proposition 29. Therefore, by definition, $abcd$ is a square, which is what was proposed. The same may be done otherwise: let $ac$ be perpendicular upon line $ab$ through Proposition 11, and let it be equal to it as before; and from point $c$, through Proposition 31, let $cd$ be drawn parallel to $ab$, and let it be placed equal to it, and let line $db$ be drawn, which through Proposition 33 will be equal and parallel to $ac$. All the angles are right angles through the last part of Proposition 29, wherefore by definition we have the proposal.

¶ Proposition 46.

In every right-angled triangle, the square which is described from the side opposite the right angle, being multiplied by itself, is equal to the two squares which are written from the two remaining sides.
This is the famous Pythagorean Theorem, though here it follows the medieval numbering as the 46th proposition of the first book.
Let $abc$ be a triangle whose angle $a$ is a right angle. I say that the square of side $bc$ is equal to the square of $ab$ and the square of $ac$ taken together. I shall square the three sides according to the teaching of the preceding [proposition]: let the square of $bc$ be the surface $bcde$, and the square of $ba$ be the surface $bfga$, and the square of $ac$ be the surface $achk$. From the right angle $a$, I shall lead to the base $de$ (the base of the last square) three lines: $fal$ parallel to each side $bd$ and $ce$, which shall cut $bc$ at point $m$, and the connecting linesoriginal: "ypothemisas". A medieval term for lines connecting the vertices of the squares in this specific proof, derived from "hypotenuse". $ad$ and $ae$. Likewise, from the two remaining angles of the triangle, which are $b$ and $c$, I shall lead to the two angles of the two smaller squares two lines intersecting each other within the same...