THE COLLECTION. BOOK V 39l 
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fits having 
tlte 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 G 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 :ABGA. C B D F 
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): (sector ABE), 
and, componendo, AFBH: {EBH) > A ABF: {ABE). 
But (by an easy lemma which has just preceded) 
A FBH: {EBH) = A ABF: {ABD), 
whence A ABF: {ABD) > A ABF: {ABE), 
and {ABE) > {ABD). 
Therefore {ABE): {ABC) > {ABD): {ABC) 
> {ABD): A ABC, a fortiori. 
Therefore Z BAF: Z BAG > {ABD): A ABC, 
whence, inversely, A ABC: {ABD) > Z BAG: Z BAF. 
and, componendo, (sector AGD): {ABD) > R: Z BCA. 
[If a be the circular measure of ¿BGA, this gives (if AG=b) 
^ odd 1 \ {^ odd 1 — sin a cos a. 6‘ 2 ) > \ ;a, 
or 2a : (2a — sin 2a) > n: 2a; 
that is, 6/{6 — sin 6) > tt/6, where 0 < 6 < tt.] 
(2) ABC is again a triangle right-angled at B. With C as 
centre and CA as radius draw a circle AD meeting BG pro 
duced in D. To prove that 
(sector CAD): (area ABD) > R: ¿AGD.