III.15 Proposition III.15
Of straight lines in a circle the diameter is greatest, and of the rest the nearer to the centre is always greater than the more remote.
Proof
Let AD be a diameter, and BC any other chord. From the centre
E drop EF⊥BC (I.12). By III.3, F is the midpoint of
BC, so BF=BC/2. By I.47 in △EBF: EB2=BF2+EF2. Since EB= radius =AD/2, (AD/2)2=(BC/2)2+EF2, hence BC2=AD2−4⋅EF2<AD2 when EF>0. So
the diameter AD is the longest chord. For two non-diameter
chords with distances EF1<EF2 from the centre, the same
identity gives the chord through F1 longer than the one through
F2.
lines 74–74 in main.tex
