328 
THE SQUARING OF THE CIRCLE 
and AK as radius, draw the quadrant KFL cutting the quad- 
ratrix in F and AB in L. 
Join AF, and produce it to meet the circumference BED 
in E] draw FH perpendicular to AD. 
Now, by hypothesis, 
(arc BED): AB = AB:AK 
= (arc BED): (arc LFK); 
therefore AB = (arc LFK). 
But, by the property of the quadra- 
trix, 
AB: FH = (arc BED): (arc ED) 
— (arc LFK): (arc FK); 
and it was proved that AB = (arc LFK); 
therefore FH = (arc FK): 
which is absurd. Therefore AK is not greater than AG. 
(2) Let AK be less than AG. 
With centre A and radius AK draw the quadrant KML. 
Draw KF at right angles to AD meeting the quadratrix 
in F; join AF, and let it meet the 
quadrants in M, E respectively. 
Then, as before, we prove that 
AB = (are LMK). 
And, by the property of the quad 
ratrix, 
AB: FK = (arc BED): (arc DE) 
' K G D = (arc LMK): (arc MK). 
Therefor^ since AB = (arc LMK), 
FK — (arc KM): 
which is absurd. Therefore AK is not less than AG. 
Since then AK is neither less nor greater than AG, it is 
equal to it, and 
(arc BED): AB = AB :AG. 
[The above proof is presumably due to Dinostratus (if not 
to Hippias himself), and, as Dinostratus was a brother of 
Menaechmus, a pupil of Eudoxus, and therefore probably