393 
PAPPUS OF ALEXANDRIA 
Draw AE at right angles to AC. With A as centre and 
AC as radius describe the circle FGE meeting AB produced 
in F and AE in E. 
Then, since ¿AGD > ¿CAE, (sector AGD) > (sector ACE). 
Therefore (AGD) : A ABC > (ACE) : AABO 
> (AGE) : (ACF), a fortiori, 
> ¿eac-.lgab. 
Inversely, 
A ABC: (ACD) < ¿GAB: ¿EAG, 
and, componcndo, 
(ABB) : (AGD) < ¿ EAB : 4 EAG. 
Inversely, (AGD) : (ABB) > Z EAG : ¿ EAB 
> E : ¿ AGB. 
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 
E 
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, AG respectively. Join BH, and draw 
LHM parallel to DF.