r

256

THE DIVISION OF

TRACT 14.

progression 1,3, 5, 7, &c ; and here again the sums of the ex-tremes, and of every two equidistant means, make up the seve-ral equal parts of the circle. Which is the other condition.”

3. But this subject admits of a more geometrical form, andis capable of being rendered very general and extensive, andis moreover very fruitful in curious consequences. For first,in whatever ratio the whole diameter is divided, whether intoequal or unequal parts, and whatever be the number of parts,the perimeters of the spaces will al-

equal, and constant, by the same,

whatever be the ratio of the parts ad, dc, cb, of the diame-ter. We shall presently find too that the spaces tv, its, andpq, will be universally as the same parts ad, dc, cb, of thediameter.

4. The semicircles having been described as before men-tioned, erect cf. perpendicular to ab, anti join be. Thenwill the circle on the diameter be, be equal to the space pq.For, join ae.

Now the space p = semicircle on ab —semicircle on ac;but the semicir. on ab = semicir. on ae + semicir. on be,and the semicir. on ac = semicir. on ae — semicir. on ce,theref. semic. ab —semic. Ac = semic. be + semicir. ce,that is, the space p is =:semic. be + semicir. ce ;

to each of these add the space ci, or the semicircle on bc,then p -p a — semic. be + semic. ce + semic. bc,that isp -p a = double the semic. be, or = the whole circleon be.

5. In like manner, the two spaces pq and rs together, orthe whole space pqrs, is equal to the circle on the diameterbf. And therefore the space rs alone, is equal to the differ-ence, or the circle on bf minus the circle on be.

ways be equal. For since the circum-ferences of circles are in the sameratio as their diameters, and becauseab and ad + db and ac + cb are allequal, therefore the semi-circumfer-ences c and b -f d and a + e are all

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