Proof
Let be the given finite straight line. With centre and
distance describe the circle (Postulate 3). With centre
and distance describe the circle (Postulate 3). From
the point , where the circles cut one another, draw and
(Postulate 1). Since is the centre of ,
(Definition I.15). Since is the centre of ,
(Definition I.15). By Common Notion 1, . Therefore the
triangle is equilateral.
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Full neighborhood
Depends on (4)
- 1Postulate 1To draw a straight line from any point to any point.
- 3Postulate 3To describe a circle with any centre and distance.
- I.15Definition I.15A circle is a plane figure contained by one line such that all the straight lines falling upon it from one point among…
- 1Common notion 1Things which are equal to the same thing are also equal to one another.
Required by (dependents) (4)
- I.2Proposition I.2To place at a given point (as an extremity) a straight line equal to a given straight line.
- I.9Proposition I.9To bisect a given rectilineal angle.
- I.10Proposition I.10To bisect a given finite straight line.
- I.11Proposition I.11To draw a straight line at right angles to a given straight line from a given point on it.
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