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. |