FGHKL is the duplicate of that which BA has to GF: therefore as the square of BM to the square of GN, so is the polygon ABCDE to the polygon FGHKL. Wherefore, similar polygons, &c. Q. E. D.

PROPOSITION II.

20.6.

THEOR.-Circles are to one another as the squares of their See N. diameters.

Let ABCD, EFGH be two circles, and BD, FH their diameters: as the square of BD to the square of FH, so shall the circle ABCD be to the circle EFGH.

For if it be not so, the square of BD n.ust be to the square of FH, as the circle ABCD is to some space either less than the circle EFGH, or greater than it ‡. First let it be to a space s less than the circle EFGH; and in the circle EFGH† † 6.4. describe the square EFGH. 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:

is less than the square described about it

and the circle * 41. 1. therefore the

square EFGH is greater than half of the circle Divide the

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: but every seg-41. 1. ment is less than the parallelogram in which it is: wherefore

+ For there is some square equal to the circle ABCD; let P be the side of it, and to three straight lines, BD, FH, and P, there can be a fourth proportional; let this be Q: therefore the squares of these four straight lines are † 22.6. proportionals; that is, to the squares of BD, FH, and the circle ABCD, it is possible there may be a fourth proportional. Let this be S. And in like manner are to be understood some things in some of the following propositions.

† 11. 11.

* 8. 11.

* 18. 11.

* 4 Def. 11. *3 Def. 11.

the planes there be taken any points, and from them perpendiculars be drawn to the planes in which the first named angles are; and from the points in which they meet the planes, straight lines be drawn to the vertices of the angles first named; these straight lines shall contain equal angles with the straight lines which are above the planes of the angles.

Let BAC, EDF be two equal plane angles: and from the points A, D, let the straight lines AG, DM be elevated above the planes of the angles, making equal angles with their sides, each to each, viz. the angle GAB equal to the angle MDE, and GAC to MDF; and in AG, DM, let any points G, M be taken, and from them, let perpendiculars GL, MN be drawn to the planes BAC, B EDF, meeting these planes in the points L, N; and join LA, ND: G the angle GAL shall be equal to the angle MDN.

F

A

D

C
KE
L

N

M

Make AH equal to DM, and through H, draw HK parallel to GL: but GL is perpendicular to the plane BAC; wherefore HK is perpendicular * to the same plane. From the points K, N, to the straight lines AB, AC, DE, DF, draw perpendiculars KB, KC, NE, NF: and join HB, BC, ME, EF. Because HK is perpendicular to the plane BAC, the plane HBK which passes through HK, is at right angles * to the plane BAC; and AB is drawn in the plane BAC at right angles to the common section BK of the two planes; therefore AB is perpendicular * to the plane HBK, and makes right angles * with every straight line meeting it in that plane: but BH meets it in that plane; therefore ABH is a right angle: for the same reason, DEM is a right angle, and is therefore equal to the angle ABH: and the angle HAB is equal + to the angle MDE; therefore in the two triangles HAB, MDE, there are two angles in one, equal to two angles in the other, each to each, and one side equal to one side, opposite to one of the equal angles in each, viz. HA equal to DM; therefore the remaining sides are equal *, each to each; wherefore AB is equal to DE. In the same manner, if HC and MF be joined, it may be demonstrated, that AC is equal to DF: therefore, since AB is equal to DE, BA and AC are equal to ED and DF, each to each; and the angle BAC is equalt to the angle EDF; where

fore the base BC is equal to the base EF, and the remaining 4.1. angles to the remaining angles; therefore the angle ABC is

equal to the angle DEF: and the right angle ABK is equal to the right angle DEN; whence the remaining angle CBK is

B

C
KE
L

N

H
G

M

in the two triangles BCK, EFN, there are two angles in one, equal to two angles in the other, each to each, and one side equal to one side adjacent to the equal angles in each, viz. BC equal to EF; therefore the other sides are equal to the other sides; BK then is equal to EN: and AB is equal to DE; wherefore AB, BK are equal to DE, EN, each to each; and they contain right angles; wherefore the base AK is equal to the base DN. And since AH is equal to DM, the square of AH is equal to the square of DM: but the squares of AK, KH are equal to the square * of AH, because AKH is a right*47.1. angle; and the squares of DN, NM are equal to the square of DM, for DNM is a right angle; wherefore the squares of AK, KH are equal to the squares of DN, NM: and of these the square of AK is equal to the square of DN; therefore the remaining square of KH is equal to the remaining square of NM; and the straight line KH to the straight line NM; and because HA, AK are equal to MD, DN, each to each, and the base HK to the base MN, as has been proved, therefore the angle HAK is equal to the plane MDN. Therefore, if from the vertices, 8.1. &c. Q. E. D.

COR. From this it is manifest, that if from the vertices of two equal plane angles, there be elevated two equal straight lines containing equal angles with the sides of the angles, each to each; the perpendiculars drawn from the extremities of the equal straight lines to the planes of the first angles, are equal to one another.

Another Demonstration of the Corollary.

Let the plane angles BAC, EDF be equal to one another, and let AH, DM be two equal straight lines above the planes of the angles, containing equal angles with BA, AC, ED, DF, each to each, viz. the angle HAB equal to MDE, and HAC equal to the angle MDF; and from H, M, let HK, MN be

* 13. 11.

perpendiculars to the planes BAC, EDF: HK shall be equal

to MN.

Because the solid angle at A is contained by the three plane angles BAC, BAH, HAC, which are, each to each, equal to the three plane angles EDF, EDM, MDF, containing the solid angle at D; the solid angles at A and D are equal, and therefore coincide with one another; to wit, if the plane angle BAC be applied to the plane angle EDF, the straight line AH coincides with DM, as was shewn in Prop. B. of this book: and because AH is equal to DM, the point H coincides with the point M: wherefore HK, which is perpendicular to the plane BAC, coincides with MN * which is perpendicular to the plane EDF, because these planes coincide with one another. Therefore HK is equal to MN. Q. E. D.

See N.

*26.11.

PROPOSITION XXXVI.

८

THEOR. If three straight lines be proportionals, the solid parallelopiped described from all three, as its sides, is equal to the equilateral parallelopiped described from the mean proportional, one of the solid angles of which is contained by three plane angles equal, each to each, to the three plane angles containing one of the solid angles of the other figure.

Let A, B, C be three proportionals, viz. A to B, as B to C: the solid described from A, B, C shall be equal to the equilateral solid described from B, equiangular to the other.

Take a solid angle D contained by three plane angles EDF, FDG, GDE; and make each of the straight lines ED, DF, DG equal to B, and complete the solid parallelopiped DH: make LK equal to A, and at the point K, in the straight line LK, make * a solid angle contained by the three plane angles LKM, MKN, NKL, equal to the angles EDF,

11

0

G

FDG, GDE, each to each; and make

KN equal to B, and KM equal to

C; and complete the solid paral-

lelopiped KO. And because, as

A is to B, so is B to C, and that A is equal to LK, and B is equal to each of the straight lines DE, DF, and C equal to KM; therefore LK is to ED as DF to KM; that is, the sides about the equal angles are reciprocally proportional; therefore the parallelogram LM is equal to EF: and because EDF, LKM 14.6. are two equal plane angles, and the two equal straight lines DG, KN are drawn from their vertices above their planes, and contain equal angles with their sides; therefore the perpendiculars from the points G, N, to the planes EDF, LKM are equal to * one another; therefore the solids KO, DH are of the same altitude; and they are upon equal bases LM, EF; 11. and therefore they are equal to one another: but the solid KO is described from the three straight lines A, B, C, and the solid DH from the straight line B. Therefore, if three straight lines, &c. q. E. D.

Cor. 35.

31.11.

PROPOSITION XXXVII.

THEOR. If four straight lines be proportionals, the similar See N. solid parallelopipeds similarly described from them shall also be proportionals: and if the similar parallelopipeds similarly described from four straight lines be proportionals, the straight lines shall be proportionals.

Let the four straight lines AB, CD, EF, GH be proportionals, viz. as AB to CD, so EF to GH; and let the similar parallelopipeds AK, CL, EM, GN be similarly described from them, AK shall be to CL, as EM to GN.

Make AB, CD, O, P continual proportionals, as also 11. 6. and because as AB is to CD so EF to GH;

EF, GH, Q, R:

and that CD is

to O, as GH to

E

FGH QR

Next let the solid AK be to the solid CL, as the solid EM to the solid GN: the straight line AB shall be to CD, as EF to GH.

Take as AB to CD, so EF to ST, and from ST describe * a solid parallelopiped SV similar and similarly situated to either