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BOOK XII.

DEFINITIONS.
I.

A PYRAMID is said to be inscribed in a cone, when its base is inscribed in the base of the cone, and both solids have a common vertex; and, a cone is said to be described about a pyramid, when its base is described about the base of the pyramid, and both solids have a common vertex.

II.

A cone is said to be inscribed in a pyramid, when its base is inscribed in the base of the pyramid, and both solids have a common vertex; and, a pyramid is said to be described about a cone, when its base is described about the base of the cone, and both solids have a common vertex.

III.

A prism is said to be inscribed in a cylinder when its bases are inscribed in the bases of the cylinder; and, a cylinder is said to be described about a prism, when its bases are described about the bases of the prism.

IV.

A cylinder is said to be inscribed in a prism when its bases are inscribed in the bases of the prism; and, a prism is said to be described about a cylinder, when its bases are described about the bases of the cylinder.

V.

A polyhedron is said to be inscribed in a sphere when the vertices of its solid angles are in the superficies of the sphere; and, a sphere is said to be described about a polyhedron when the superficies of the sphere passes through the vertices of its solid angles.

VI.

A sphere is said to be inscribed in a polyhedron when its superficies touches the faces or planes of the polyhedron; and, a polyhedron is said to be described about a sphere, when its faces or planes touch the superficies of the sphere.

POSTULATES.
I.

Let it be granted that a square or other rectilineal figure may contain the same area or space as a circle; or, that to two squares and a circle there may be a fourth proportional.

II.

Let it be granted that to two triangles and a pyramid erected on one of them, or to two similar rectilineal figures and a solid erected on one of them, there may be a fourth proportional; and that to any three solids there may be a fourth proportional.

The preceding definitions and postulates are not laid down in Euclid's Elements. But as they are understood and taken for granted in this Book, they are formally inserted here for the benefit of the learner.

The following Lemma is the first proposition of the Tenth Book of Euclid's Elements, and is usually inserted here, as being necessary to the understanding of the demonstration of some of the propositions of this Book.

LEMMA L

If from the greater of two unequal magnitudes of the same kind, there be taken more than its half, and from the remainder more than its half; and so on: there will at length remain a magnitude less than the least of the proposed magnitudes.

Let A B and C be two unequal magnitudes, of the same kind, of which AB is the greater. If from A B there be taken more than its half, and from the remainder more than its half, and so on; there will at length remain a magnitude less than C.

K

For C may be multiplied so as at length to become greater than A B. Let it be so multiplied, and let D E its multiple be greater than AB; also let DE be divided into parts DF, FG, and GE, each equal to C. From AB take B H greater than its half, and from the remainder AH take HK greater than its half, and so on, until there be as H many divisions in AB as there are in DE: and let the divisions in A B be AK, KH, and H B; and the divisions in DE be DF, FG, and G E.

Because DÉ is greater than AB, and E G taken from

D

DE is not greater than its half, but BH taken from A B is greater than its half. Therefore the remainder G D is greater than the remainder HA. Again, because GD is greater than HA, and that GF is not greater than the half of G D, but HK is greater than the half of HA. Therefore the remainder F D is greater than the remainder A K. But FD is equal to C. Therefore C is greater than AK; that is, A K is less than C. Wherefore, if from the greater, &c. Q.E.D.

COROLLARY.-If only the halves be taken away, the same thing may in the same way be demonstrated.

PROP. I. THEOREM.

Similar polygons inscribed in circles, are to one another as the squares of their diameters.

Let ABCDE and FGHKL be two circles having the similar polygons ABCDE and FG HK L inscribed in them; and let B M and GN be the diameters of the circles. The square of B M is to the square of GN, as the polygon ABCDE is to the polygon F GHKL.

Join BE, AM, GL, and FN.

Because the polygon ABCDE is similar to the polygon FG HK L, the

angle BAE is equal to the angle B

GFL, and BA is to A E as G F is to

N

FL. Therefore the two triangles

M

BAE and GFL (VI. 6) are equi

angular; and the angle A E B is equal

to the angle FLG. But the angle A E B is equal (III. 21) to the angle

A M B, because they stand upon the same arc.

For the same reason, the

angle FLG is equal to the angle FNG. Therefore also the angle AMB is equal to the angle FNG; and the angle BA M is equal to the (III. 31) angle GFN. Therefore the remaining angles in the triangles A B M and FGN are equal, and they are equiangular to one another. Wherefore B M is to GN, as (VI. 4) BA is to G F. And the duplicate ratio of BM to GN, is the same (V. 5 and 22, Def. 10) with the duplicate ratio of BA to G F. But the ratio of the square of B M to the square of GN, is the duplicate (VI. 20) ratio of that which BM has to GN; and the ratio of the polygon ABCDE to the polygon F GHKL is the duplicate (VI. 20) of that which B A has to G F. Therefore the square of BM is to the square of G N, as the polygon A B C D E is to the polygon FGHKL. Wherefore, similar polygons, &c. Q. E. D. Corollary-Similar polygons are to one another as the squares of the radii of their inscribed or circumscribed circles, or as the squares of the chords of similar segments.

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Circles are to one another as the squares of their diameters.

Let A B C D and E F G H be two circles, and BD and FH their diameters. The square of B D is to the square of F H as the circle A B C D is to the circle E F G H.

For, if not, the square of BD is to the square of FH, as the circle ABCD is to some space either less than the circle EFGH, or greater than it (XII. Post. 1). First, let it be to a space S less than the circle EFGH; and in the eircle E F G H (IV. 6) describe the square EFGH. This square is greater than half of the circle EFGH; because, if through the points E, F, G, and H, tangents be drawn to the circle, the square EFGH is half (I. 41) of the square described about the circle. But the circle is less than the square described about it. Therefore the square EFGH is greater than half of the circle. Bisect the arcs EF, FG, GH, and II E, at the points K, L, M, and N, and join E K, KF, FL, LG, G M, M H, H N, and N E. Each of the triangles EK F and FL G is greater than half of the segment which contains it. For if straight lines touching the circle be drawn through the points K, L, M, and N, and the parallelograms upon the straight lines EF, FG, GH, and HE be completed, each of the triangles E KF, FL G, GM H, and HNE is the half (1. 41) of the parallelogram which contains it. But every segment is less than the parallelogram which contains it. Therefore each of the triangles EKF, FLG, GM H, and HN E is greater than half the segment which contains it. Again, if the arcs EK, KF, &c. be bisected, and their extremities be joined; and so on: there will at length remain segments of the circle, which taken together are less than the excess of the circle E F G H above the space S. For (XII. Lemma 1) if from the greater of two unequal magnitudes there be taken more than its half, and from the

X

remainder more than its half, B

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the least of the proposed magni

A

P

c

E

K

G

tudes. Let the segments EK, KF, FL, LG, GM, MH, HN, and NE be those which 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 EK FLGMHN. Because the square of BD is to the square of FH (XII. 1) as the polygon AXBOCPDR is to the polygon EKFLG MÍN. But the square of BD is to the square of FH (Hyp.) as the circle ABCD is to the space S. Therefore the circle ABCD is to the space S, as (V. 11) the polygon AXBOCPDR is to the polyglon EKFLGMH N. But the circle A B CD is greater than the polygon contained in it. Therefore the space S is greater (V. 14) than the polygon EK FLGMHN; but it is also less, as has been proved, which is impossible. Therefore the square of BD is not to the square of FH, as the circle ABCD is to any space less than the circle EFGH. In the same manner, it may be demonstrated, that the square of FH is not to the square of B D, as the circle EFGH is to any space less than the circle A B CD. Neither is the square of BD to the square of FH, as the circle A B CD to any space greater than the circle EFG H. For, if possible, let the square of BD be to the square of FH, as the circle ABCD is to T, a space greater than the circle EFGH. Therefore, inversely, the square of FH is to the square of B1), as the space T is to the circle ABCD. But the space T is to the circle ABCD), as the circle E FGH is to some space, which must be_less (V. 14) than the circle ABCD, because the space T is greater. (Hyp.) than the circle E F G H. Therefore

the square of FH is to the

A

N

square of BD, as the circle EFGH is to a space less than the circle A B C D, which has been proved to be impossible. Therefore the square of BD is not to the square of FH as the circle ABCD is to any space greater than the circle E F G H. But it has been proved, that the square of BD is not to the square of FH, as the circle A B CD is to any space less than the circle E F G H. Therefore, the square of BD is to the square of FH, as the circle A B CD is to the circle E F G H. Therefore, circles are, &c. Q. E. D.

Corollary 1.-Similar polygons are to one another as their inscribed or circumscribed circles.

Corollary 2.-Circles are to one another, as the squares of their radii, or as the squares of the chords of similar segments; and so are these segments.

PROP. III. THEOREM.

Every triangular pyramid may be divided into two equal and similar triangular pyramids similar to the whole pyramid, and two equal prisms which are together greater than half of the whole pyramid. Let ABCD be a pyramid of which the base is the triangle B C and

10s vertex the point D. The pyramid may be divided into two equal and similar triangular pyramids similar to the whole pyramid, and two equal prisms which are together greater than half of the whole pyramid.

Bisect AB, BC, CA, AD, DB, and D C, at the points E, F, G, H, K and L; and join EH, EG, GH, HK, KL, LH, EK, KF, FG, and E F.

B

D

H

K

E

G

F

Because A E is equal to E B, and AH to HD, HE HK is is parallel (VI. 2) to D B. For the same reason, parallel to A B. Therefore HE BK is a parallelogram, and HK is equal (I. 34) to E B. But EB is equal to AE. Therefore also AE is equal to HK, and AH is equal to HD. Wherefore EA and AH are equal to KH and HD, each to each; and the angle EAH is equal (I. 29) to the angle K HD. Therefore the base EH is equal to the base KD, and the triangle A EH is equal (I. 4) and similar to the triangle HKD. For the same reason, the triangle A G H is equal ana similar to the triangle HLD. Again, because EH and HG, which meet one another, are parallel to KD and DL, that meet one another, but are not in the same plane with them, they contain equal (XI. 10) angles. Therefore the angle EHG is equal to the angle K D L. Because EH and H G are equal to KD and DL, each to each, and the angle EHG equal to the angle KDL. Therefore the base EG is equal to the base KL, and the triangle E HG equal (I. 4) and similar to the triangle KD L. For the same reason, the triangle A E G is also equal and similar to the triangle HK L. Therefore the pyramid AEGH is equal (XI. C) and similar to the pyramid K HL D. Because HK is parallel to AB, a side of the triangle AD B, the triangle ADB is equiangular to the triangle HD K, and their sides are (VI. 4) proportionals. Therefore the triangle ADB is similar to the triangle HDK. For the same reason, the triangle D B C is similar to the triangle D KL; the triangle ADC to the triangle HDL; and the triangle ABC to the triangle A EG. But the triangle AEG was proved to be similar to the triangle HKL. Therefore the triangle ABC is similar (VI. 21) to the triangle H KL. Wherefore the pyramid ABCD is similar (XI. B, and XI. Def. 11) to the pyramid HKLD. But the pyramid HKLD is similar, as has been proved, to the pyramid AEG H. Therefore the pyramid ABCD is similar to the pyramid AEGH. Wherefore each of the pyramids AEGH and HKLD is similar to the whole pyramid ABCD. Because BF is equal to FC, the parallelogram EBFG is double (I. 41) of the triangle GF C. But the prism EBFGHK having the parallelogram EBF G for its base, is equal (XI. 40) to the prism G F CHKL, having the triangle G F C for its base; for they are of the same altitude, viz., the distance between the parallel (XI. 15) planes A B C and H KL. But each of these prisms is greater than either of the pyramids AEGH and HKLD. For the prism EBFGHK on the base EBFG, is greater than the pyramid E B FK, on the base E BF. But the pyramid EBFK is equal (XI. C) to the pyramid AEG H; because they are contained by equal and similar planes. Therefore the prism EBFGHK is greater than the pyramid AEGH. But the prism EBF GHK is

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