as the circle ABCD is to some space either less than the circle EFGH, or greater than it.

First, if possible, let it be to a space Sless than a circle EFGHI; and in the circle EFGH inscribe the square EFGH. (IV. 6.) This square is greater than half of the circle EFGH;

because, if through the points E, F, G, H, there be drawn tangents to the circle,

the square EFGH is half of the square described about the circle: (1.47.) and the circle is less than the square described about it;

therefore the square EFGH is greater than half the circle. Divide the circumferences EF, FG, GH, HE, each into two equal parts in the points K, L, M, N, and join EK, KF, FL, LG, ĠM, HM, HN, NE;

therefore each of the triangles EKF, FLG, GMH, HNE, is greater than half of the segment of the circle in which it stands;

because, if straight lines touching the circle be drawn through the points K, L, M, N, and the parallelograms upon the straight lines EF, FG, GH, HE be completed,

each of the triangles EKF, FLG, GMH, HNE is the half of the parallelogram in which it is: (I. 41.)

but every segment is less than the parallelogram in which it is; wherefore each of the triangles EKF, FLG, GMH, HNE is greater than half the segment of the circle which contains it.

Again, if the remaining circumferences be divided each into two equal parts, and their extremities be joined by straight lines, by continuing to do this, there will at length remain segments of the circle, which together are less than the excess of the circle EFGH above the space S;

because, by the preceding Lemma, if from the greater of two unequal magnitudes there be taken more than its half, and from the remainder more than its half, and so on, there shall at length remain a magnitude less than the least of the proposed magnitudes.

Let then the segments EK, KF, FL, LG, GM, MH, HN, NE be those that remain, and are together less than the excess of the circle EFGH above S:

therefore the rest of the circle, viz. the polygon EKFLGMHN is greater than the space S.

Describe likewise in the circle ABCD the polygon AXBOCPDR similar to the polygon EKFLGMHN:

as therefore the square on BD is to the square on FH, so is the polygon AXBOCPDR to the polygon EKFLGMHN: (XII. 1.)

but the square on BD is also to the square on FH, as the circle ABCD is to the space S; (hyp.)

therefore as the circle ABCD is to the space S, so is the polygon AXBOCPDR to the polygon EKFLĠMHN: (v. 11.)

but the circle ABCD is greater than the polygon contained in it; wherefore the space Sis greater than the polygon EKFLGMHN: (v. 14.)

but it is likewise less, as has been demonstrated; which is impossible. Therefore the square on BD is not to the square on FH, as the circle ABCD is to any space less than the circle ÊFGH.

In the same manner, it may be demonstrated, that neither is the square on FH to the square on BD, as the circle EFGH is to any space less than the circle ABCD.

Nor is the square on BD to the square on FH, as the circle ABCD is to any space greater than the circle EFGH.

For, if possible, let it be so to T, a space greater than the circle EFGH;

therefore, inversely, as the square on FH to the square on BU, so is the space T to the circle ABCD;

but as the space Tis to the circle ABCD, so is the circle EFGH to

some space, which must be less than the circle ABCD, (v. 14.) because the space T'is greater, by hypothesis, than the circle EFGH therefore as the square on FH is to the square on BD, so is the circle EFGH to a space less than the circle ABCD, which has been demonstrated to be impossible;

therefore the square on BD is not to the square on FH as the circle ABCD is to any space greater that the circle EFGH:

and it has been demonstrated, that neither is the square on BD to the square on FH, as the circle ABCD to any space less than the circle EFGH:

wherefore, as the square on BD is to the square on FH, so is the circle ABCD to the circle EFGH.

NOTES TO BOOK XIL

THE first comparison of rectilinear areas is made in the first book of the Elements by the principle of superposition, where two triangles are coincident in all respects; next, comparison is made between triangles and other rectilinear figures when they are not coincident.

In the sixth book, similar triangles are compared by shewing that they are in the duplicate ratio of their homologous sides, and then by dividing similar polygons into the same number of similar triangles, and shewing that the polygons are also in the duplicate ratio of any of their homologous sides.

In the eleventh book, similar rectilinear solids are compared by shewing that their volumes are to one another in the triplicate ratio of their homologous sides.

In the twelfth book a new principle is introduced, called "the method of exhaustions," which is founded on the principle of exhausting a magnitude or the difference of two magnitudes, by successively taking away a certain part of it.

The method of exhaustions was employed by the Ancient Geometers and was strictly rigorous in its principles; but it was too tedious and operose in its application to be of extensive utility as an instrument of investigation. It is exemplified in Euc. XII. 2, where it is proved that the areas of circles are proportional to the squares on their diameters. In demonstrating this truth, it is first shewn by inscribing successively in one of the circles, regular polygons of four, eight, sixteen, &c. sides, and thus tending to exhaust the difference between the areas of the circle and polygon, that a polygon may be found which shall differ from the circle by an area less than any magnitude that can be assigned: and then since similar polygons inscribed in circles are as the squares on their diameters (Euc. XII. 1), the truth of the proposition is established by means of an indirect proof.

"The method of exhaustions" may be applied to find the circumference and area of a circle. A rectilineal figure may be inscribed in the circle and a similar one circumscribed about it, and then by continually doubling the number of sides of the inscribed and circumscribed polygons, by this principle, it may be demonstrated, that the area of the circle is less than the area of the circumscribed polygon, but greater than the area of the inscribed polygon; and that as the number of sides of the polygon is increased, and consequently the magnitude of each diminished, the differences between the circle and the inscribed and circumscribed polygons are continually exhausted.

In a similar way the principle is applied to the volumes and surfaces of the sphere, cone, &c.

The Second Proposition of the twelfth book is perhaps retained merely as an example of the method employed by the Ancient Geometers. This method has been replaced by the method of prime and ultimate ratios, which is now employed in the proofs of such propositions as were formerly effected by the method of exhaustions.

GEOMETRICAL EXERCISES ON BOOK XII.

THEOREM I.

If semicircles ADB, BEC be described on the sides AB, BC of a rightangled triangle, and on the hypotenuse another semicircle AFBGC be described, passing through the vertex B; the lunes AFBD and BGCE are together equal to the triangle ABC.

B

E

C

It has been demonstrated (XII. 2) that the areas of circles are to one another as the squares on their diameters; it follows also that semicircles will be to each other in the same proportion.

Therefore the semicircle ADB is to the semicircle ABC, as the square on AB is to the square on AC,

and the semicircle CEB is to the semicircle ABC, as the square on BC is to the square on AC,

hence the semicircles ADB, CEB, are to the semicircle ABC as the squares on AB, BC are to the square on AC;

but the squares on AB, BC are equal to the square on AC, (1. 47.) therefore the semicircles ADB, CEB are equal to the semicircle ABC. (v. 14.)

From these equals take the segments AFB, BGC of the semicircle on AC, and the remainders are equal,

that is, the lunes AFBD, BGCE are equal to the triangle BAC.

THEOREM II.

If on any two segments of the diameter of a semicircle, semicircles be described, all towards the same parts, the area included between the three circumferences (called apßnλos) will be equal to the area of a circle, the diameter of which is a mean proportional between the segments.

Let ABC be a semicircle whose diameter is AB,
and let AB be divided into any two parts in D,

B

and on AD, DC let two semicircles be described on the same side; also let DB be drawn perpendicular to AC.

Then the area contained between the three semicircles, is equal to the area of the circle whose diameter is BD.

Since AC is divided into two parts in C,

the square on AC is equal to the squares on AD, DC, and twice the rectangle AD, DC; (II. 4.)

and since BD is a mean proportional between AD, DO;

the rectangle AD, DC is equal to the square on DB, (vi. 17.) therefore the square on AC is equal to the squares on AD, DC, and twice the square on DB.

But circles are to one another as the squares on their diameters or radii, (XII. 2.)

therefore the circle whose diameter is A Cis equal to the circles whose diameters are AD, DC, and double the circle whose diameter is BD; wherefore the semicircle whose diameter is AC is equal to the circle whose diameter is BD, together with the two semicircles whose diameters are AD and DC:

if the two semicircles whose diameters are AD and DC be taken from these equals,

therefore the figure comprised between the three semi-circumferences is equal to the circle whose diameter is DB.

THEOREM III.

There can be only five regular solids.

If the faces be equilateral triangles. The angle of an equilateral triangle is one-third of two right angles; and six angles, each equal to the angle of an equilateral triangle, are equal to four right angles: and therefore a number of such angles less than six, but not less than three are necessary to form a solid angle. Hence there cannot be more than three regular figures whose faces are equal and equilateral triangles.

If the faces be squares. Since four angles, each equal to a right angle, can fill up space round a point in a plane. A solid angle may be formed with three right angles, but not with a number greater or less than three. Hence, there cannot be more than one regular solid figure whose faces are equal squares.

If the faces be equal and regular pentagons. Since each angle of a regular pentagon is a right angle and a fifth of a right angle: the magnitude of three such angles being less than four right angles, may form a solid angle, but four, or more than four, cannot form a solid angle. Hence, there cannot be more than one regular figure whose faces are equal and regular pentagons.

If the faces be equal and regular hexagons, heptagons, octagons, or any other regular figures; it may be shewn that no number of them can form a solid angle.

Wherefore there cannot be more than five regular solid figures, of which, there are three, whose faces are equal and equilateral triangles; one, whose faces are equal squares; and one, whose faces are equal and regular pentagons.