If the method employed in the text for discovering the radius of the circle, which has twice the number of sides under the same extent of perimeter, be applied to the hexagon or its elementary equilateral triangle, the numbers will stand as below.

Wherefore, .95492965855:1:: 3:3.1415926536; and hence 3.1415926536 is the nearest expression, consisting of ten decimal places, for the area of the circle whose radius is 1. But the semicircumference in this case denoting also the surface, the same number must represent the circumference of a circle whose diameter is 1. Consequently, if D denote the diameter of any circle, the circumference will be expressed approximately, by 3.1415926536xD; whence the area will be Dx 3.1415926536, or Dx.7853981634.

By help of the note to Prop. 27. Book V. lower numbers may be found, approximating to the same results. For in this case

m3, n=7, p=16, and q=11: whence, remounting from these conditional equalities, the ratio of the diameter to the circumference of a circle is denoted progressively, by 1: 3-by 7: 22-by 113: 355—and by 1250 : 3927. The ratio of 1 to 3 is the rudest approximation, being the same as that of the diameter of the circle to the perimeter of its inscribed hexagon; the ratio of 7 to 22 is what was discovered by Archimedes; the ratio of 113 to 355, in which the three first odd numbers appear in pairs, was first proposed by Adrian Metius of Alkmaer, Professor of Mathematics and Medicine at Franeker, who died in 1636; and the ratio of 1250 to 3927, the same as 1 to 3.1416, is that generally adopted by the Hindus. Hence also the circle is to its circumscribing square nearly as 11 to 14, or, still more nearly-as 355 to 452.

To this Book may be added the following Propositions.

PROP. I. THEOR.

If from any point in the circumference of a circle, straight lines be drawn to the extremities of a chord, and meeting the perpendicular diameter, they will divide that diameter, internally and externally, in the same ratio.

Let the chord EF be perpendicular to the diameter AB of a circle, and from its extremities F and E straight lines FG and EG be inflected to a point G in the circumference, and cutting the diameter internally and externally in C and D ; then will AC: CB:: AD :: DB.

For join AG and BG, and draw HBI parallel to AG.

Because AEGB is a semicircle, the angle AGB is a right angle (III. 19.); wherefore AG and HI being parallel, the alternate angle GBI is right (I. 22.), and likewise its adjacent angle GBH. But the diameter AB, being perpendicular to the chord

EF, must (III. 4. and 13.) bisect the arc FAE, and therefore the angle EGA is equal to AGF (III. 12. cor.) or (III. 17.), its supplement. And since AG is parallel to HI, the angle EGA is equal to the angle GIB or its supplement (1. 22.); and for the same reason, the an

gle AGF is equal to the alternate angle GHB. Whence the angle GIB is equal to GHB; but the angles GBI and GBH being both right angles, are equal, and the side GB is common to the two triangles BIG and BHG, which are, therefore, equal (I. 20.), and consequently BH is e

E

G

qual to BI, and AG : BH : : AG : BI. Now, because the pa rallels AG and BH are intercepted by the diverging lines AB and GH, AG: BH :: AC: CB (VI. 2.); and since the parallels AG and BI are intercepted by the diverging lines GD and AD, AG: BI:: AD: DB. Wherefore, by identity of ratios, AC CB:: AD: DB, that is, the straight line AB is cut in the same ratio, internally and externally, or the whole line AD is divided harmonically in the points C and B.

:

Cor. 1. As the points E and G come nearer each other, it is obvious that the straight line EGD will approach continually to the position of the tangent, which is its ultimate limit. Hence the tan-.

A

E

gent and the perpendicular, from the point of contact or mutual coincidence, cut the diameter proportionally, or AC : CB: AD: DB. It is, therefore, evident (VI. 7.) that, O being the centre, OC: OB:: OB: OD.

Cor. 2. Since OC: OB:: OB: OD, it follows (V. 19. cor. 2.) that OC: OD :: OB2-OC or AC.CB: OD2 — OB2 or AD.DB; whence, by division, CD: OD:: AD.DB-AC.CB, or VI. 7. cor.) CD2; AD.DB.

PROP. II. THEOR.

If two straight lines be inflected from the extremities of the base of a triangle to cut the opposite sides proportionally, another straight line, drawn from the vertex through their point of concourse, will bisect the base.

In the triangle ABC, let AE and CD, drawn from the extremities of the base to cut the opposite sides proportionally, intersect each other in F, join BF, which produce if necessary to meet the base in the point G; AG will be equal to GC, For join DE. And because the sides AB and BC are cut proportionally, DE is parallel to AC

H

D

E

(VI. 1. cor.), whence BD: BA :: BH: BG (VI. 1.); but BD : BA :: DE: AC (VI. 2.), and therefore BH: BG:: DE: AC. Again, the parallels DE and AC being cut by the diverging lines AE and CD, DE: AC: DF: FC (VI. 2.) and DF: FC:: FH; FG (VI. 1.); wherefore BH: BG :: FH: FG, or BF is cut internally and externally in the same ratio. But DH being parallel to AG, BH: BG:: DH: AG; and since DH is also parallel to GC, HF: FG:: DH: GC; whence DH; AG:: DH: GC, and consequently AG is equal to GC,

:

PROP. III. THEOR.

G

C

If a semicircle be described on the side of a rectangle, and through its extremities two straight lines be drawn from any point

in the circumference to meet the opposite side produced both ways; the altitude of the rectangle will be a mean proportional between the segments thus intercepted.

Let ABED be a rectangle, which has a semicircle ACB described on the side AB, and the straight lines CA and CB drawn from a point C in the circumference to meet the extension of the opposite side DE; the altitude AD of the rectangle will be a mean proportional between the exterior segments FD and EG.

For, the angle ADF, being evidently a right angle, is equal to the angle ACB, which stands in a semicircle (III. 19.). and

the angle DFA is equal to the exterior angle BAC (I. 22.); wherefore (VI. 11.) the triangle FAD is similar to ABC. In the same manner, it is proved that the triangle BGE is similar to ABC; whence the triangles DAF and BGE are similar to each other, and consequently (VI. 11.) FD: AD :: BE or AD: EG.

If the straight lines CD and CE be drawn, they will (VI. 2.) divide the diameter AB into segments AH, HI, and IB, which are respectively proportional to the segments FD, DE, and EG of the extended side DE. Consequently when ABED is a square, and therefore DE a mean proportional between FD and EG, it must follow that HI is likewise a mean proportional between AH and IB.

If the rectangle ABED have its altitude AD equal to the side of a square inscribed within the circle, the square of the