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Source — Euclid's Elements, encoded as an rrxiv paper — rrxiv

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Source — Euclid's Elements, encoded as an rrxiv paper

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I.1 Proposition I.1

On a given finite straight line to construct an equilateral triangle.

Proof

Let

A B

be the given finite straight line. With centre

A

and distance

A B

describe the circle

B C D

(Postulate 3). With centre

B

and distance

B A

describe the circle

A C E

(Postulate 3). From the point

C

, where the circles cut one another, draw

C A

and

C B

(Postulate 1). Since

A

is the centre of

B C D

,

A C = A B

(Definition I.15). Since

B

is the centre of

A C E

,

B C = B A

(Definition I.15). By Common Notion 1,

A C = B C

. Therefore the triangle

A B C

is equilateral.

Figure

Proposition I.1. The two circles with centres $A$ and $B$ and
common radius $AB$ meet at $C$; the triangle $ABC$ is equilateral. — Proposition I.1. The two circles with centres $A$ and $B$ and common radius $A B$ meet at $C$ ; the triangle $A B C$ is equilateral.

lines 74–74 in main.tex