[ 435 1 
the radius, as the fquare of F G, or the redangle 
EFC, to the redangle under AG (or AC) and FB, 
But, EB being to B F as BF to BC, by converfio.n, 
E B is to E F as B F to F C, and alfo, by taking the 
difference of the antecedents and of the confequents, 
EF is to twice AF as BF to FC ; and twice AFB 
is equal, to E F C. 
Now, let the triangle BAH be formed, where 
the angle B A H is greater than BAG. Flere, the 
perpendicular H I being drawn, the redangle under 
the fine of B A H and the tangent of A B H will be 
to the fquare of the radius, as the redangle EIC to 
the re&ahgle under AC, IB. But IF is to FB 
as 2 A F I to 2 A F B, or EFC; and 2 A F I 
is greater than AF? — A I?; alfo AF? — A I? to- 
gether with EFC, is equal to EIC; therefore, by 
compofition, the ratio of I B to BF is greater than 
that of E I C to EFC; and the ratio of AC x I B 
to A C x F B greater than that of E I C to EFC: 
alfo, by permutation, the ratio of ACxJB to EIC 
greater than the ratio of A C x F B to E FC. But 
the fir ft of thefe ratios is the fame with that of the 
fquare of the radius to the re dangle under the fme of 
BAH and the tangent of ABH; and the latter is 
the fame with that of the fquare of the radius to the 
redangle under the fine of B A G and the tangent 
of ABG ; therefore, the latter of thefe tw r o redangles 
is greater than the other. 
Again, let the triangle BAK be formed, with the 
angle BAK lets than BAG, and the perpendicular 
K L be drawn. Then the redangle under the fine of 
BAK and the tangent of ABK is to the fquare of 
the radius, as the fquare of KL to the redangle under 
Kkk 2 AC, 
