The Elements

...of the equilateral triangle This refers to the triangle constructed in the first part of Proposition 2 on the previous page. which is opposite to the given point, namely side D.C., I shall extend through the center of the described circle as far as its circumference; and let the whole line thus extended be D.C.E., according to the length of which I will describe a circle with the center at D, which shall be circle E.F. After this, I shall extend side D.A. as far as the circumference of this last circle, and let it meet its circumference at point F. I say, therefore, that A.F. is equal to B.C. For B.C. and C.E. are equal, because they extend from the center of circle C.B. to its circumference. Similarly, D.F. and D.E. are also equal because they extend from the center of circle E.F. to the circumference. But D.A. and D.C. are equal because they are sides of an equilateral triangle. Therefore, if the equal parts D.A. and D.C. are taken away from D.E. and D.F. (which are also equal), the remaining parts A.F. and C.E. will be equal. Since both lines A.F. and C.B. are equal to C.E., they are equal to each other. Therefore, from point A, we have drawn line A.F. equal to B.C., which is what was proposed.

A geometric diagram showing two intersecting circles and a triangle. A line d-c-e extends from the center d of the larger circle through the center c of the smaller circle. A triangle a-b-d is constructed within the area where the circles overlap.

A diagram showing two parallel lines, a-b and c-d. A circle is centered on point c, with a radius equal to line a-b, intersecting line c-d at point f.

Given two unequal lines, to cut off from the longer of them a segment equal to the shorter.
Let there be two lines, A.B. and C.D., and let A.B. be the shorter. I wish to cut from C.D. a line that is equal to A.B. I first draw from point C a single line equal to A.B., using the method taught in the preceding proposition; let this be C.E. Then, placing the center at point C, I shall describe a circle according to the length of C.E., which will intersect line C.D. Let it intersect it at point F. Line C.F. will be equal to line C.E. because both extend from the center of the same circle to its circumference. And since both lines A.B. and F.C. are equal to C.E., they are equal to each other; which is what was proposed.

Two sets of diagrams. The first shows two identical triangles, a-b-c and d-e-f. The second shows a complex triangular construction with points a, b, c, e, f, g, and h, used to illustrate the superposition proof for congruent triangles.

In all cases of two triangles where two sides of the one are equal to two sides of the other, and the two angles contained by those equal sides are equal to one another: the remaining sides corresponding to each other will also be equal, the remaining angles of the one will be equal to the remaining angles of the other, and the whole triangle will be equal to the whole triangle.
Let there be two triangles A.B.C. and D.E.F., and let side A.B. be equal to side D.E., and side A.C. equal to side D.F., and angle A equal to angle D. Then I say that the base B.C. is equal to base E.F., and angle B equal to angle E, and likewise angle C equal to angle F. This is proven as follows: I will superimpose triangle A.B.C. upon triangle D.E.F. so that angle A falls upon angle D, side A.B. upon side D.E., and side A.C. upon side D.F. It is clear through the penultimate common notion original: "penultimā pceptionē"; this refers to Euclid's Common Notion 4: "Things which coincide with one another are equal to one another." that neither the angles nor the sides will exceed one another, because angle A is equal to angle D, and the sides are placed upon those to which they are equal by hypothesis. Therefore, points B and C will fall upon points E and F. If line B.C. falls exactly upon line E.F., the proposition is clear; because when line B.C. is superimposed on line E.F. and neither exceeds the other, they are equal by the converse of the penultimate common notion. By the same reasoning, angle B will be equal to angle E, and angle C equal to angle F. However, if line B.C. does not fall upon line E.F., but falls inside the triangle (like line E.G.F.) or outside (like line E.H.F.), then two straight lines would enclose a surface, which is against the final postulate original: "ultimā petitionē"; in early editions, the principle that two straight lines cannot enclose a space was often listed as a postulate or common notion..

A diagram of an isosceles triangle a-b-c. The sides a-b and a-c are extended downwards to points d and e. Lines cross from b to e and from c to d, forming a larger triangular structure beneath the base b-c.

In every triangle with two equal sides an isosceles triangle, it is necessary that the angles at the base be equal to each other; and if the two equal sides are extended, the two angles below the base will also be equal to each other.
Let there be triangle A.B.C. whose side A.B. is equal to side A.C. I say that angle A.B.C. is equal to angle A.C.B.; and if A.B. and A.C. are extended as far as D and E...