construct equal segments by extension

I.2

To place at a given point (as an extremity) a straight line equal to a given straight line.

—Euclid

From the given points \(A\) and \(B\) set segment \(\bar{AB}\).

From the given points \(C\) and \(D\) set segment \(\bar{CD}\) shorter than \(\bar{AB}\).

PROBLEM: Extend segment \(\bar{CD}\) to a length equal to \(\bar{AB}\)

construct circle \((A, B)\) as \(c_1\)

With points \(A\) and \(C\) construct segment \(\bar{AC}\)

contruct equilateral triangle on \(\bar{AC}\)

identify new point of equaliteral triangle as \(E\)

construct line \(\overline{EA}\)

construct line \(\overline{EC}\)

Identify new point on \(c_1\) intersecting with \(\overline{EA}\) as \(F\)

construct circle \((E, F)`\) as \(c_2\)

Identify new point on \(c_2\) intersecting with \(\overline{EC}\) as \(G\)

Let the straight lines AE, BF be produced [2] in a straight line with DA, DB; [I.post.2] with centre B and distance BC let the circle CGH be described; [I.post.3] and again, with centre D and distance DG let the circle GKL be described. [I.post.3]

Then, since the point B is the centre of the circle CGH,

\(\therefore\) BC is equal to BG.

Again, since the point D is the centre of the circle GKL,

\(\therefore\) DL is equal to DG.

And in these DA is equal to DB;

\(\therefore\) the remainder AL is equal to the remainder BG [3].

[I.c.n.3]

But BC was also proved equal to BG;

\(\therefore\) each of the straight lines AL, BC is equal to BG.

And things which are equal to the same thing are also equal to one another; [I.c.n.1]

\(\therefore\) AL is also equal to BC.

\(\therefore\) at the given point A the straight line AL is placed equal to the given straight line BC.

• (Being) what it was required to do.

dependencies

[I.def.15]: /elem.1.def.15 “Book I - Definition 15” [I.1]: /elem.1.1 “Book I - Proposition 1” [I.post.1]: /elem.1.post.1 “Book I - Postulate 1” [I.post.2]: /elem.1.post.2 “Book I - Postulate 2” [I.post.3]: /elem.1.post.3 “Book I - Postulate 3” [I.c.n.1]: /elem.1.c.n.1 “Book I - Common Notion 1” [I.c.n.3]: /elem.1.c.n.3 “Book I - Common Notion 3”

footnotes

[1]

(as an extremity).

I have inserted these words because <quote>to place a straight line <em>at</em> a given point</quote> (<foreign lang=”greek”>πρὸς τῷ δοθέντι σημείῳ</foreign>) is not quite clear enough, at least in English.

[2]

Let the straight lines AE, BF be produced….

It will be observed that in this first application of <a href=”/elem.1.post.2”>Postulate 2</a>, and again in <a href=”/elem.1.5”>I. 5</a>, Euclid speaks of the <em>continuation</em> of the straight line as that which is produced in such cases, <foreign lang=”greek”>ἐκβεβλήσθωσαν</foreign> and <foreign lang=”greek”>προσεκβεβλήσθωσαν</foreign> meaning little more than <em>drawing</em> straight lines <quote>in a straight line with</quote> the given straight lines. The first place in which Euclid uses phraseology exactly corresponding to ours when <pb n=”245”/>speaking of a straight line being produced is in <a href=”/elem.1.16”>I. 16</a>: <quote>let one side of it, BC, be produced to D </quote> (<foreign lang=”greek”>προσεκβεβλήσθω αὐτοῦ μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ</foreign>).

[3]

the remainder AL…the remainder BG. The Greek expressions are <foreign lang=”greek”>λοιπὴ ἡ ΑΛ</foreign> and <foreign lang=”greek”>λοιπῇ τῇ</foreign> BH, and the literal translation would be <quote>**AL** (or BG) <em>remaining</em>,</quote> but the shade of meaning conveyed by the position of the definite article can hardly be expressed in English.