parallelopiped described from A, B, C as its edges shall be equal to the equilateral parallelopiped described from B, as its common edge, equiangular to the other.

Take a solid angle D contained by three plane angles EDF, FDG, and GDE; and make each of the straight lines ED, DF, and DG equal to B, and complete the parallelopiped D H. Take any straight line LK equal to A, and at the point K in the straight line LK, make (XI. 26) a solid angle contained by the three plane angles L K M, MKN and NK L, equal to the angles EDF, FDG, and GD E, each to each; make KN equal to B, and KM equal to C; and complete the parallelopiped KO.

L

N

H

Because, A is to B, as B is to C, and A is equal to LK, and B' to each of the straight lines DE, DF, and C to KM. Therefore LK is to ED, as DE is to KM; that is, the sides about the equal angles are reciprocally proportional; therefore the parallelo gram LM is equal (VI. 14) to the parallelogram EF. Because EDF and LKM are two equal plane angles, and the two equal straight lines DG and K N are drawn from their vertices above their planes; and contain equal angles with their sides. Therefore the perpendiculars from the points G and N to the planes E D F and LK M are equal to (XI. 35 Cor.) one another. Wherefore the solids K O and DH are of the same altitude: and they are upon equal bases LM, and EF. Therefore the parallelopiped KO is equal (XI. 31) to the parallelopiped DH; and the solid K O is described from the three straight lines A, B, and C, as its edges, and the solid DH from the straight line B as its common edge. Therefore if three straight lines, &c. Q. E. D.

Exercise. Demonstrate the converse of this proposition.

PROP. XXXVII, THEOREM.

B

If four straight lines be proportionals, the similar parallelopipeds similarly described from them as edges are also proportionals; and, conversely, if the similar parallelopipeds similarly described from four straight lines as edges be proportionals, the straight lines are proportionals. Let the four straight lines A B, CD, EF and GH, be proportionals, viz., AB to CD, as EF to GH; and let the similar parallelopipeds AK, CL, E M and GN be similarly described from them as edges. AK is to CL, as EM is to GN.

100

Make (VI. 11) AB, CD, O and P continual proportionals, as also EF, GH, Q and R.

K

Because A B is to CD, as EF to GH; and CD is (V. 11) to O, as GH is to Q; and O is to P, as Q is to R. Therefore, ex æquali (V. 22) AB is to P, as E F is to R. But A B is to P, as (XI. Cor. 33) the solid AK, is to the solid CL; and EF is to R, as (XI. Cor. 33) the solid EM is to the solid GN. There

B C

I

fore (V. 11) the solid AK is to the solid CL, as the solid E M is to the solid GN.

Next, let the solid AK be to the solid CL, as the solid EM to the solid G N. The straight lines AB, CD, EF, and GH, are proportionals.

F G H

Q

R

Take A B to CD, so EF to ST, and E from ST as edge, describe (XI. 27)a parallelopiped S V similar and similarly situated to either of the parallelopipeds, EM or G N.

Because A B is to CD, as E F is to ST, and that from A B and CD as edges the parallelopipeds AK and CL are similarly described; and the parallelopipeds EM and SV from the straight lines EF and ST as edges. Therefore A K is to CL, as EM is to S V. But (Hyp.) A K is to CL, as EM is to G N. Therefore G N is equal (V. 9) to S V. But it is similar and similarly situated to S V. Therefore the planes which contain the solids GN and SV are similar and equal, and their homologous sides G H and ST equal to one another. Because A B is to CD, as EF is to ST, and ST is equal to GH. Therefore AB is to CD, as EF is to GH. Therefore, if four straight lines, &c. Q. E. D.

PROP. XXXVIII. THEOREM.

If one plane be perpendicular to another, a straight line drawn from a point in the one perpendicular to the other, meets their common section.

Let the plane CD be perpendicular to the plane A B, and let AD be their common section; if any point E be taken in the plane CD, the perpendicular, drawn from E to the plane A B, meets A D.

From E draw EG in the plane CD perpendicular to A D (I. 12) and in the plane AB, from the point G, draw GF perpendicular to AD (I. 11).

Because EG is perpendicular to GF (XI. Def. 3), and also to AD (Const.) Therefore E G is C perpendicular to the plane AB (XI. 4). But from the point E, no other straight line can be drawn perpendicular to A B, than EG (XI. 13). Therefore the perpendicular drawn from E to the plane AB meets AD. Wherefore, if one plane be pendicular, &c. Q. E. D.

1

per

E

B

Dr. Simson considers this proposition as an interpolation in this Book, and quite out of place.

In a parallelopiped, if the sides of two of the opposite planes be each bisected, the common section of the planes passing through the points of bisection and any diagonal of the parallelopiped, bisect each other.

Let the sides of the opposite planes CF and A H, of the parallelopiped AF, be bisected in the points K, L, M, and N; and X, O, P, and R. Join KL, MN, XO, and PR. Because DK and CL are equal and parallel, KL is parallel (I. 33) to DC. For the same reason, M N is parallel to BA. But BA is parallel to DC. Because KL and BA are each of them parallel to DC, and not in the same plane with it, KL is parallel (XI. 9) to BA; and because KL and MN are each of them parallel to

D

K

F

0

E

BA, and not in the same plane with it, KL is parallel (XI. 9) to MN. Therefore KL and MN are in one plane. In like manner it may be proved, that XO 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

Р

N

Because DX is parallel to OE, the alternate angles DXY and YO E are equal (I. 29). Because D X is equal to OE and X Y to Y O, 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 BG join their extremities. Therefore DE 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 in T. 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., DY to GS; for they are the halves of DE and B G. Therefore the remaining sides are equal (I. 26), each to each. Wherefore D T is equal to T G, and Y T to T S. Therefore, if in a parallelopiped, &c. Q. E.D.

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.

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 NG; 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 AX and GO. Because the parallelogram AF is

double of the triangle GHK ; and

B

E

D

M

Χ

N

H

F

G K

the parallelogram HK double (I. 34) of the same triangle. Therefore the parallelogram AF is equal to HK. But parallelopipeds upon equal bases, and of the same altitude, are equal(XI. 31) to one another. Therefore the solid AX 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) o the solid GO. Therefore the prism A BCDEF is equal to the prism GHKLMN. Wherefore, if two triangular prisms, &c. Q. E. D.

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 bere, 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 haif; 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 BHK 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 AB be AK, KH, and H B; and the divisions in DE be DF, FG, and GE.

B

D

Because DÉ is greater than AB, and E G 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 H K is greater than the half of HA. Therefore the remainder F D is greater than the remainder AK. But FD is equal to C. Therefore C is greater than AK; that is, AK 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 GHK L.

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
FL. Therefore the two triangles
BAE and GFL (VI. 6) are equi-
angular; and the angle A E B is equal

E G

N

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