A history of Greek mathematics - Wilbourhall.org

THE COLLECTION. BOOK V 391
the same circumference the semicircle is the greatest, with some
preliminary lemmas which deserve notice (chaps. 15, 16).
(1) ABC is a triangle right-angled at B. With C as centre
and radius CA describe the arc
AD cutting CB produced in D.
To prove that (R denoting a right
angle)
(sector CAD) :
(area ABD)
> R./BCA.
Draw AF at right angles to CA meeting CD produced in F,
and draw BH perpendicular to AF. With A as centre and
AB as radius describe the arc GBE.
«
Now (area EBF) : (area EBH) > (area EBF) :
and, componendo, AFBH: (EBH) > AABF: (ABE).
But (by an easy lemma which has just preceded)
whence
and
Therefore (ABE) :
AFBH: (EBH) = AABF: (ABD),
AABF: (ABD) > AABF: (ABE),
(ABE) > (ABD).
(ABG) > (ABD) : (ABG)
(sector ABE),
> (ABD): A ABC, a fortiori.
Therefore Z BAF: Z BA C > (ABD) : A ABC,
whence, inversely,
and, componendo, (sector ACD) :
[If a.
AABC:(ABD) > Z BAG: Z BAF.
(ABD) > R : Z BCA.
be the circular measure of /.BCA, this gives (if AC=b)
%otb 2 :(^ocb — 2 -J
sin a cos a . 6 2 ) >^7r:a,
or
2a:(2a — sin2a) > tt:2oc;
that is, 0/(0 — sin 0) > n/0, where < < tt.]
centre and CA as radius draw a circle
(2) ABC is again a triangle right-angled at B. With C as
AD meeting BC produced
in D.
To prove that
(sector CAD) :
(area ABD) > R :
/.ACD.