The Elements

... (faded marginal note in cursive script) ...?

...and of the remaining sides, those by the center; the angles contained by the sides bg and ad, neither are the right angles coincident in the same way, [nor] the sides ab and ag. I say, therefore, that the sides ba and ag are greater than the remaining side bg... [as is] evident from bg, if any... ...? which was to be demonstrated.

... (marginal note) ...?

If two straight lines are constructed within a triangle from the endpoints of one of the sides, the lines so constructed will be less than the remaining two sides of the triangle, but will contain a greater angle. For let there be a triangle abg, and from the endpoints of the side bg, let two straight lines bd and dg be constructed within; I say that the lines bd and dg are less than the remaining two sides of the triangle, ba and ag, but that they contain a greater angle, bdg, than the angle bag. For let bd be extended to e. And since in any triangle the two sides are greater than the remaining one, the two sides ba and ae of the triangle abe are greater than the remaining side be. Let the common segment eg be added; therefore, ba and ag are greater than be and eg. Again, since the two sides ge and ed of the triangle gde are greater than the remaining side gd, let the common segment db be added; therefore, ge and eb are greater than gd and db. But ba and ag were shown to be greater than be and eg; therefore, by much, ba and ag are greater than bd and dg. Again, since in any triangle the exterior angle is greater than the interior and opposite angle, the exterior angle bdg of the triangle gde is greater than the interior angle ged. For the same reasons, therefore, the exterior angle geb of the triangle abe is greater than the interior angle bag. But bdg was shown to be greater than geb; therefore, by much, bdg is greater than bag. If, therefore, two straight lines are constructed within a triangle from the endpoints of one of its sides, the lines so constructed will be less than the remaining two sides of the triangle, but will contain a greater angle; which was to be demonstrated.

A geometric diagram showing a triangle with vertices A, B, and Γ. Inside the triangle is a point Δ. Lines are drawn from vertices B and Γ to point Δ. The line from B to Δ is extended to intersect the side AΓ at point E. This illustrates Euclid's Elements, Book I, Proposition 21.

... (notes in right margin) ...?

Let the given finite straight line be ab. It is required, then, to construct an equilateral triangle on the straight line ab. With center a and distance ab, let the circle bgd be drawn; and again, with center b and distance ba, let the circle age be drawn; and from the common point g, where the circles cut each other, let the straight lines ga and gb be joined to the points a and b. And since point a is the center of the circle bgd, ag is equal to ab. Again, since point b is the center of the circle age, bg is equal to ba. But ag was also shown to be equal to ab; therefore, each of the lines ga and gb is equal to ab. And things equal to the same thing are also equal to each other; therefore, ga is also equal to gb; therefore, the three lines ga, ab, and bg are equal to each other. Therefore, the triangle abg is equilateral; and it has been constructed upon the given finite straight line ab; which was to be done.

A geometric diagram for Euclid's Proposition I.1. It shows two intersecting circles with centers labeled A and B. The centers are connected by a line segment, and both are connected to one of the intersection points, labeled Γ, forming an equilateral triangle. There are also labels Δ and Ε on the circumferences.

To place a straight line equal to a given straight line at a given point.

Let the given point be a, and the given straight line be bg; it is required, then, to place a straight line equal to the given straight line bg at the point a. For let the straight line ab be joined from point a to point b; and on it let an equilateral triangle dab be constructed, and let the straight lines ae and bz be extended in a straight line with da and db; and with center b and distance bg, let the circle ghth original: "ΓΗΘ" be drawn; and again, with center d and distance dh, let the circle hkl original: "ΗΚΛ" be drawn. Since, therefore, point b is the center of the circle ghth, bg is equal to bh; again, since point d is the center of the circle hkl, dl is equal to dh; of which da is equal to db. Therefore, the remainder al is equal to bh. But bg was also shown to be equal to bh; therefore, each of al and bg is equal to bh. And things equal to the same thing are also equal to each other; therefore, al is also equal to bg. Therefore, a straight line al equal to the given straight line bg is placed at the given point a; which was to be done.

A more complex geometric diagram illustrating Euclid's Proposition I.2. It features two large intersecting circles, one centered at B and a larger one centered at Δ. An equilateral triangle ΔΑΒ is shown with sides extended through points A and B to the circumferences of the circles at points Λ and H. To the right of the circles are three vertical line segments labeled Γ, Ε, Δ.