halves of the other pair: these four halves therefore are equal to two right angles; and if they be taken from four right angles, the sum of the angles of the triangles ABE, CDG, there remain the angles E, G equal to two right angles; and therefore (APP. I. 4.) a circle may be described about the quadrilateral EFGH.

When the circles are described internally, the proof proceeds on the same general principle, but is rather more simple.*

PROP. XXXV. PROB.

To divide a given straight line into two parts, such that their rectangle may be equal to a given space, not. exceeding (II. 5. cor. 2.) the square of half the given line.

Let AB be a given straight line: it is required to divide it into two parts, such that their rectangle may be equal to a given space, not greater than the square of the half of AB.

Bisect AB in C: then if the given space be equal to the square of AC or CB, AB is divided in C in the manner required. But if not, on AB describe a semicircle; draw ED parallel to AB, and at a distance equal to the

D

side of a square containing the given space; and draw DG perpendicular to AB: G is the point of section required.

For (VI. 13.) DG is a mean proportional between AG, GB, and (VÌ. 17.) the rectangle AG.GB is equal to the square of GD; that is (const.) to the given space.

PROP. XXXVI. PROB.

To produce a given straight line, so that the rectangle under the whole produced line, and the part produced, may be equal to a given space.

Let it be required to produce a given straight line AB, so that

* If in this case, the circle which touches three of the sides, touches also the fourth, the four circles, as also their centres, may be regarded as coinciding; and the circle passing through the four centres, is reduced to a point.

the rectangle under the whole produced line, and the part added, may be equal to the square of a given straight line C.

Bisect AB in D, and draw BE perpendicular to AB, and equal to C: from D as centre, at the distance DE, describe a semicircle cutting AB produced in F and G: either of these is the extremity of the produced part.

E

B F

For (VI. 13.) BE is a mean proportional between GB, BF, or AF, BF, since AF, GB are obviously equal. Therefore (VI. 17.) the rectangle AF.FB is equal to the square of BE, that is, of C.

Schol. This proposition and the foregoing are the most useful cases of the 28th and 29th of the sixth book.

PROP. XXXVII. THEOR.

A LINE ABC, whether curved or polygonal, which is convex throughout, that is, which can be cut by a straight line in only two points, is less than any line, whether curved or polygonal, which envelops it, from one end to the other.

For, if ABC be not shorter than any of the lines that envelop it, some one of those lines will be less than any of the others, and will be either less than ABC, or, at most, equal to it. Let AHFGEC be that line, and draw the straight line DG cutting it in D, G, and either touching ABC, or not meeting it. Then DG is less than DHFG, and consequently (I. ax. 4.) ADGEC is less than AHFGEC: but, by hypothesis, AHFGEC is less than any other enveloping

F

G

E

B

D

line; and yet the enveloping line ADGEC is less than it, which is absurd: therefore every enveloping line is greater than ABC.

Schol. In exactly the same manner it may be shown, that a convex line returning into itself, is shorter than any

line enveloping it on all sides, and either touching it, or encompassing it without touching it.

PROP. XXXVIII. THEOR.

THE circumferences of circles are to each other as their radii.

Let AB, CD be the radii of two circles; their circumferences are proportional to these.

For if the circumference of the circle OR be not to that of FK, as AB to CD, AB will be to CD, as the circumference of OR to the circumference of a circle having a radius either less or greater than CD; and first, suppose its radius to be CE less than CD. In the circle FK inscribe a regular polygon, the sides of which will not meet the circle EN, and in OR describe a similar polygon. *

Then, if AT, AO, CM, CF were joined, we should have (VI. 4.) TO: MF:: AT or AB : CM or CD. But (VI. 20. cor. 4.) the perimeter of the polygon in OR is to that of the polygon in FK, as TO to MF, that is, as AB to CD: and, by hypothesis, AB is to CD, as the circumfer

E

O

ence of OR to the circumference of EN: therefore (V. 11.) the perimeter of the polygon in OR is to that of the polygon in FK, as the circumference of OR to the circumference of EN, which (V. 14.) is absurd; as, by the preceding proposition, the first term of this analogy is less than the third, and the second greater than the fourth. Hence AB cannot be to CD, as the circumference of OR to a circumference less than that of FK: and it would be proved in a similar manner, that CD cannot be to AB, as the circumference of FK to a circumference less than that of OR.

Neither can AB be to CD, as the circumference of OR to one greater than that of FK. For if it could, we should have, by inversion, CD to AB, as a circumference greater than that of FK to the circumference of OR, or, which is equivalent, as the circumference of FK to one less than that of OR. But this has been proved to be impossible: wherefore AB is to CD, as the circumference of OR to the circumference of FK.

Cor. Hence the arcs of sectors which have equal angles, are as the radii of the sectors. For the arcs are evidently proportional to the entire circumferences, the angles at the centres being equal.

PROP. XXXIX. THEOR.

THE area of a circle is equal to the rectangle under its radius, and a straight line equal to half its circumference.

Let AB be the radius of the circle BC: the area of BC is equal

The method of effecting the construction here pointed out, is evident from the 16th proposition of the twelfth book.

to the rectangle under AB, and a straight line D equal to half the circumference.

G

H

B

M

For, if the rectangle AB.D be not equal to the circle BC, it is equal to a circle either greater or less than BC. First, suppose, if possible, the rectangle AB.D to be the area of a circle EF, of which the radius AE is greater than AB, and let GHK be a regular polygon described about the circle BC, such that its sides do not meet the circumference of EF. Then, by dividing this polygon into triangles by radii drawn to G, H, K, &c., it would be seen that its area is equal to the rectangle under AB and half its perimeter. But (APP. I. 37.) the perimeter of the polygon is greater than the circumference of AB: and therefore the area of the polygon is greater than the rectangle AB.D, that is, by hypothesis, than the area of the circle EF, which is absurd, since the polygon is only a part of that circle, being contained within it. The rectangle AB.D, therefore, is not equal to the area of any circle greater than BC.

D

In the second place, this rectangle cannot be less than BC. For suppose, if possible, that it is equal to the area of a circle described from A as centre, within BC. Then, by describing about that circle a regular polygon, the sides of which would not meet BC, it would be shown as before, that the area of the polygon would be equal to the rectangle under half its perimeter and the radius of the interior circle, and would therefore be less than the rectangle AB.D, because AB is greater than the radius, and (APP. I. 37.) D greater than half the perimeter. Therefore the polygon being less than AB.D, would be less than the circle about which it is described, which is absurd.

Hence the rectangle under the radius AB and half the circumference, is not equal to a circle greater than BC, nor to one less : it is therefore equal to BC.

Cor. 1. Hence the surface of a sector ALM is equal to the rectangle under its radius and half its arc. For the circle BC is to ALM, as the circumference of BC to the arc LM, sectors and arcs being proportional (VI. 33.) to their angles at the centre. Hence, by halving the third and fourth terms, BC is to ALM, as D to LM, or as D.AB to LM.AB: therefore (V. 14.) LM.AB is equal to the sector ALM, because BC is equal to

D.AB.

Cor. 2. Let the circumference of the circle whose diameter is unity, and consequently half the circumference of the circle whose radius is unity, be denoted by 7. Then (APP. I. 38.) 1 is to T, as 2AB to the circumference of BC. Hence we find the circum

ference to be 2 X AB; and multiplying half of this by the radius AB, we find the area to be equal to X AB2. Hence the area of any circle is found by multiplying the square of its radius by the constant number, which, as will be shown in the scholium to the next proposition, is 3.141593 nearly.

PROP. XL. PROB.

THE area of a regular inscribed polygon, and that of a regular circumscribed one of the same number of sides being given; to find the areas of the regular inscribed and circumscribed polygons having double the number of sides.

H

G

K

C

Let A be the centre of the circle, BC a side of the inscribed polygon, and DE parallel to BC, a side of the circumscribed one. Draw the perpendicular AFG, and the tangents BH, CK, and join BG: then BG will be a side of the inscribed polygon of double the number of sides; and (IV. B.) HK is a side of the similar circumscribed one. Now, as a like construction would be made at each of the remaining angles of the polygon, it will be sufficient to con

sider the part here represented, as the triangles connected with it are evidently to each other, as the polygons of which they are parts. For the sake of brevity, then, let P denote the polygon whose side is BC, and P' that whose side is DE; and, in like manner, let A and A', represent those whose sides are BG and HK: P and P', therefore, are given; Q and Q' required.

Now (VI. 1.) the triangles ABF, ABG are proportional to their bases AF, AG, as they are also to the polygons P, Q; therefore AF AG :: P: Q. The triangles ABG, ADG are likewise as their bases AB, AD, or (VI. 4.) as AF, AG; and they are also as the polygons Q and P': therefore AF: AG :: Q: P'; wherefore (V. 11.) P: Q:: Q: P'; so that Q is a mean proportional between the given polygons P, P'; and, representing them by numbers, we have QPP, and the area of Q will be computed by multiplying P by P', and extracting the square root of the pro

duct.

Again, because AH bisects the angle GAD, and because the triangles AHD, AHG are as their bases, we have GH: HD::AG: AD, or AF: AB:: AHG: AHD. But we have already seen that