D

K F

0

BA, and not in the same plane with it, KL is parallel (XI. 9) to M N Therefore KL and MN are in one plane. In like manner it may be proved, that X O and PR are in one plane. Let YS be the common section of the planes KN and XR; and DG the diagonal of the parallelopiped A F. The straight lines Y S and DG meet and bisect each other. Join D Y, YE, B S, and S G.

B

N

G

Because DX is parallel to OE, the alternate angles D X Y and YO E are equal (I. 29). Because D X is equal to O E and X Y to Y Ó, and they contain equal angles, the base D Y is equal (I. 4) to the base YE, and the other angles are equal. Therefore the angle XYD is equal to the angle OYE, and DYE is a straight (I. 14) line. For the same reason BSG is a straight line, and B S is equal to S G. Because CA is equal and parallel to DB, and also to E G. Therefore D B is equal and parallel (XI. 9) to EG; and DE and B G join their extremities. Therefore D'E is equal and parallel (I. 33) to BG. But DG and Y S drawn joining them are in one plane. Therefore DG and YS must meet one another; let them meet Because DE is parallel to B G, the alternate angles EDT and BGT are (XI. 29) equal. But the angle D T Y is equal (I. 15) to the angle GTS. Wherefore in the triangles DTY and GTS there are two angles in the one equal to two angles in the other, and one side equal to one side, opposite to equal angles in each, viz., D Y to GS; for they are the halves of D E and B G. Therefore the remaining sides are equal (I. 26), each to each. Wherefore DT is equal to T G, and Y T to ↑ S. Therefore, if in a parallelopiped, &c. Q. E.D.

in T.

PROP. XL. THEOREM.

If two triangular prisms of the same altitude, have the base of one a parallelo. gram, and the base of the other a triangle; and if the parallelogram be double of the triangle, the prisms are equal.

B

M

Let the prisms ABCDEF and GHKLMN be of the same altitude, of which the first is contained by the two triangles ABE and CDF, and the three parallelograms A D, DE, and EC; and the other by the two triangles GHK and LMN, and the three parallelograms LH, HN, and N G ; and let one of them have a parallelogram A F, and the other a triangle GHK, for its base: if the parallelogram AF be double of the triangle GHK, the prism ABCDEF is equal to the prism GHKLMN, Complete the solid parallelopipeds A X and GO. Because the parallelogram AF is double of the triangle GHK; and the parallelogram HK double (I. 34) of the same triangle. Therefore the parallelogram AF is equal to H K. But parallelopipeds upon equal bases, and of the same altitude, are equal(XI. 31) to one another. Therefore the solid A X is equal to the solid GO. But the prism ABCDEF is half (XI. 28) of the solid AX; and the prism GHKLMN half (XI. 28) of the solid GO. Therefore the prism ABCDEF is equal to the prism GHKLMN. Wherefore, if two triangular prisms, &c. Q. E. D.

A

E

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X

H

F

G K

N

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 I.

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 A B 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.

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 G E, each equal to C. From A B take B H greater than its half, and from the remainder A H take HK greater than its half, and so on, until there be as many divisions in AB as there are in DE: and let the divisions in A B be AK, K H, and H B; and the divisions in DE be DF, F G, and G E.

K

H

B

D

E

Because D E is greater than AB, and EG taken from DE is not greater than its half, but B H 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 H A, and that GF is not greater than the half of G D, but HK is greater than the half of H A. 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 e 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 GHK L. Join BE, AM, GL, and FN. Because the polygon ABCDE is similar to the polygon F G H K L, the angle BAE is equal to the angle GFL, and BA is to AE as G F is to F L. Therefore the two triangles BAE and GFL (VI. 6) are equiangular; and the angle A E B is equal

B

M

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 aíso the angle A M B is equal to the angle FN G; and the angle B A M is equal to the (III. 31) angle G F N. Therefore the remaining angles in the triangles A BM and F G N 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 B M to G N, 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 G N, is the duplicate (VI. 20) ratio of that which BM has to GN; and the ratio of the polygon A B CDE to the polygon F G H K L is the duplicate (VI. 20) of that which B A has to G F. Therefore the square of B M 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.

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 CD is to the circle E F G H.

For, if not, the square of BD is to the square of FH, as the circle A B C D is to some space either less than the circle E F G H, or greater than it (XII. Post. 1). First, let it be to a space S less than the circle EFGH; and in the circle E F G H (IV. 6) describe the square E F G H. 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 H E, at the points K, L, M, and N, and join E K, KF, FL, LG, G M, M H, HN, and NE. 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, G H, and HE be completed, each of the triangles E K F, FL G, G M H, and H NE 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, G M H, and HNE 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 remainder more than its half, and so on, there will at length remain a magnitude less than the least of the proposed magni

A

X

Ꭱ

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M

tudes. Let the segments E K, 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 EK FLGMHN 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 C D 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 A B CD is to any space less than the circle EFG H. In the same manner, it may be demonstrated, that the square of FH is not to the square of B D, as the circle E F G H is to any space less than the circle ABCD. Neither is the square of BD to the square of FH, as the circle A B C D 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 EFG H. Therefore,

inversely, the square of FH is to the square of BD, 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 EFGH. Therefore the square of FH is to the

A

T

square of B D, as the circle E F G H 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 A B CD 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 C D is to any space less than the circle EFGH. 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 circum scribed 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.

ABCD be a pyramid of which the base is the triangle A B C and