III.4 Proposition III.4
If in a circle two straight lines cut one another which are not through the centre, they do not bisect one another.
Proof
Let AB, CD be two chords intersecting at E, neither through
the centre F. Suppose for contradiction that E bisects both:
AE=EB and CE=ED. Join FE. By III.3 applied to chord
AB (since F is the centre and FE bisects AB at E), FE⊥AB. Applied to CD, the same line FE is ⊥CD. But
the perpendicular from F to a line is unique (I.11), so AB and
CD must coincide — contradiction with their being two distinct
chords.
lines 74–74 in main.tex
