III.1 Proposition III.1
To find the centre of a given circle.
Proof
Let ABC be the given circle. Draw any chord AB in it (Postulate
1) and bisect AB at D (I.10). From D draw DC at right
angles to AB (I.11), produced to meet the circle at C and E.
Bisect CE at F (I.10); then F is the centre. For if any other
point G were the centre, then by SSS (I.8) on △GAD and
△GBD we would obtain ∠GDA=∠GDB, both
right (I.13). But F already lies on the perpendicular bisector
of AB, and the perpendicular at D is unique (I.11); applying
the same reasoning to chord CE forces F onto its perpendicular
bisector as well. The two perpendicular bisectors meet only at the
true centre, which is F.
lines 74–74 in main.tex
