A history of Greek mathematics - Wilbourhall.org

39.2 PAPPUS OF ALEXANDRIA
Draw AE at right angles to AC. With A as centre and
AC as radius describe the circle FCE meeting AB produced
in Fand AE in E.
Then, since I ACD > ICAE, (sector ACD) > (sector AGE).
Therefore (ACD) :
A ABC > (ACE) :
Inversely,
> (ACE) : (ACF),
AABC
> LEAC-.LCAB.
a fortiori,
AABC: (ACD) < lCAB:lEAC,
Inversely, (A CD) :
and, componendo,
(ABD) : (ACD)
< Z EAB :
(ABD) > I EAC: /.EAB
>R:LACD.
Z
EAC.
We come now to the application of these lemmas to the
proposition comparing the area of a semicircle with that of
other segments of equal circumference (chaps. 17, 18).
A semicircle is the greatest of all segments of circles which
have the same circumference.
Let ABC be a semicircle with centre G, and DEF another
segment of a circle such that the circumference DEF is equal
to the circumference ABC. I say that the area of ABC is
greater than the area of DEF.
Let H be the centre of the circle DEF. Draw EHK, BG at
right angles to DF, AC respectively. Join DH, and draw
LHM parallel to DF.