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Given segment AB, we will prove that we can construct an equilateral triangle on a given finite straight line.

Draw circle with center A and radius B. | Postulate 3 |

Draw circle with center B and radius A. | Postulate 3 |

Let C be an intersection of these circles. | Implicit assumption that 2 circles in the same plane intersect. |

Connect AC. | Postulate 1 |

Connect BC. | Postulate 1 |

Notice that AC=AB. | Definition 15 and the fact that AC and AB are radii of the circle with center A. |

Notice that BC=BA. | Definition 15 and the fact that BC and BA are radii of the circle with center B. |

Hence AC=AB=BC. | Common Notion 1 and the fact that AC=AB and BC=AB. |

Therefore, ABC is equilateral. | Definition 20. |

Hence we have constructed an equilateral triangle on a given finite straight line and so we have proven proposition 1.