Step | Figure | Justification |

1 | | Start with the completed construction. I say that the circle ABC is a circle that passes through points A, B, and C. |

2 | | By the definition of a perpendicular bisector, the perpendicular bisectors PM and QM divide segments AB and BC into two equal parts and are perpendicular to AB and BC. |

3 | | Construct segments AM, BM, and CM. See Euclid. Elements. Book 1. Proposition 1. To draw a straight line from any point to any point. |

4 | | Since AP=BP AND ∠APM=∠BPM, and PM is in common, by angle-side-angle congruence, ΔAPM=ΔBPM. |

5 | | Since ΔAPM=ΔBPM, line segment AM=BM. |

6 | | Since BQ=CQ AND ∠BQM=∠CQM, and QM is in common, by angle-side-angle congruence, ΔBQM=ΔCQM. |

7 | | Since ΔBQM=ΔCQM, line segment BM=CM. |

8 | | By common notion 1, since AM=BM and BM=CM, then AM=CM. Since AM=BM=CM, points A, B, and C are equidistant from M. So a circle with center at point M and radius AM passes through B and C. Q.E.D.. |

Table 2: Proof of construction. |