But now let the centre x of the circle be in one fide MN of the triangle, and join XL: I say again that A B is greater than LX. For if it be not, A B is either equal to LX, or less than it. Firft let it be equal: Then the two fides A B, BC, that is, DE, EF are equal to the two fides MX, XL, that is, to MN. But MN is put equal to DF: Therefore DE, EF are equal to DF: which [by 20. 1.] cannot be Therefore A B is not equal to LX; nor like! wife is it lefs: For then a greater abfurdity would follow. Therefore A B is greater than LX. And if after the fame manner as above, R x be drawn at right angles to the plane of the circle fuch that its fquare be equal to the excess of the square of A B above the fquare of Rx [by the fubjoined lemma] the problem will be refolved.

B

H

Again let the centre x of the circle fall without the triangle L M N, and join LX, MX, Nx. I fay in this cafe A B is greater than LX. For if it be not, it is either equal to it, or less. First let it be equal: Then the two fides A B, BC are equal to the two fides MX, XL, each to each, and the bafe A is also equal to the base ML: Therefore [by 8. 1.] the angle ABC is equal to the angle MXL. By the fame reason the angle GHK is alfo equal to the angle LXN. Wherefore the whole angle MXN is equal to the two angles ABC, GHK. But the angles A B C, GHK are greater than the angle DEF; and therefore the angle MXN is greater than DEF. And because the two fides DE, E F are equal to the two fides MX, X N, and the bafe DF is equal to the base MN; [by 8. 1.] the angle MXN will be equal to the angle DEF. But it has been alfo proved to be greater: which is abfurd: therefore A B is not equal to LX. We shall prefently demonftrate that it

E

D

R

L

M

P

R

S

X

F

N

is not lefs: wherefore it must neceffarily be greater. And if again XR be drawn at right angles to the plane of the circle fuch, [by the fubjoined lemma] that its fquare be equal to the excefs whereby the fquare of A B exceeds the fquare of LX, the problem is refolved. Now I fay that AB is not less than LX. For if it be poffible, let it be lefs, and make xo equal to AB, and XP equal to BC, and join OP. Then because A B is equal to BC; xo will be equal to x P: Therefore the remainder oL is equal to the remainder PM: Wherefore [by 2. 6.] L M is parallel to Po; and fo the triangle LMX is equiangular to the triangle Pxo: Wherefore [by 4. 6.] as Lx is to L M, so is xo to OP: and alternately as Lx is to xo, fo is L M to OP. But LX is greater than xo; therefore L M also is greater than o P. But [by conftr.] L M is equal to AC; wherefore AC will be greater than OP; and so because the two fides A B, BC are equal to the two fides ox, XP, each to each, and the bafe AC is greater than the base OP; [by 25. 1.] the angle ABC will be greater than the angle oxp. In like manner if x R be taken equal to xo, or XP, and OR be joined, we demonftrate that the angle GHK is greater than the angle OxR. At the right line Lx, and at the point x in it, make the angle Lxs equal to the angle ABC, and the angle L x T equal to the angle G H K; make sx, XT each equal to ox, and join os, OT, ST. Then because the two fides A B, BC are equal to the two fides ox, xs, and the angle ABC is equal to the angle oxs, the bafe LC or L M [by 4. 1.] will be equal to the base o s. By the fame reafon L N is alfo equal to o T. And because the two fides ML, L N are equal to the two fides o s, OT; and the angle M L N is greater than the angles SOT; the bafe MN [by 24. 1.] is greater than the base s T. But MN is equal to DF: Therefore D F will be greater than ST. Therefore because the two fides D E, E F are equal to the two fides s x, xT, and the base DF is greater than the base s T, [by 25. 1.] the angle DEF is greater than the angle sx T. But the angle sxT is equal to the angles ABC, GHK: Therefore the angle DEF is greater than the angles ABC, GHK: but it is less too; which is impoffible,

LEMMA.

We thus demonftrate how RX may be fo taken, that its fquare fhall be equal to the excefs of the fquare of A B, above LX.

Let the right lines be AB, LX, and let AB be the greater. Upon AB defcribe the femicircle A B C: And in ABC apply the right line AC equal to L x, and join B C.

L

X A

C

B

Then because the angle A CB is in the femicircle A B C, the angle A CB [by 31. 3.] will be a right angle. Therefore the square of AB [by 47. 1.] is equal to the fquare of AC, and to the fquare of CB: Wherefore the fquare of A B exceeds the fquare of AC by the fquare of C B. But AC is equal to LX: Wherefore the fquare of AB exceeds the fquare of LX by the fquare of C B. Wherefore if XR be taken equal to C B, the fquare of AB will exceed the fquare of Lx by the fquare XR. Which was the thing to be done.

• As the measure or relative quantity of a plane right lined angle, is the fector of a circle contained under the fides of the angle, or the arch of a circle defcribed about the angular point contained between the fides of the angle. So in like manner, is the measure of a folid angle, the fector of a sphere included by the plane angles, of which that folid angle confifts; or, the furface of a sphere formed by the interfections of the plane angles of which a folid angle confifts, all meeting at the centre of the fphere, is the measure of a folid angle.

I have often thought that the theory of folid angles may be much farther extended, and perhaps made ufeful.

PROP. XXIV. THEOR.

If a folid be contained under parallel planes; the oppofite planes of it are equal, and parallelograms?.

For let the folid CDGH be contained under the parallel planes AC, GF, AH, DF, FB, AE: I fay its oppofite planes are equal and parallelograms.

For

For because the two parallel planes BG, CE are cut by the plane AC, their common fections [by 16. 11.] are parallel: Therefore A B is parallel to DC. Again, because the two parallel planes BF, A E are cut by the plane a C, their common fections are parallel: Therefore A D is parallel to BC. And because A B has been proved to be parallel to DC: Therefore AC is a parallelogram. In like manner we demonftrate that DF, FG, GB, BF, AE are each of them parallelograms,

Join A H, DF: Then because AB is parallel to DC, and BH to C F, the two right lines A B, BH touching one another will be parallel to the two right lines DC, CF touching one another, but not in the fame plane: Wherefore [by 10. 11.] they will contain equal angles: Therefore

B

H

G

C

F

the angle ABH is equal to the angle DCF. And because [by 34 1.] the two fides AB, BH are equal to the two fides DC, CF, and the angle ABH is equal to the angle DCF; the base AH [by 4. 1.] will be equal to the base DF, and the triangle ABH equal to the triangle DCF. And [by 34. 1.] the parallelogram BG is double to the triangle AB H, and the parallelogram CE to the triangle DCF: Therefore the parallelogram BG is equal to the parallelogram C E. In like manner we demonftrate that the parallelogram AC is equal to the parallelogram GF, and the parallelogram AE equal to the parallelogram BF.

If therefore a folid be contained under parallel planes, its oppofite planes will be equal and parallelograms. Which was to be demonftrated.,

P This propofition does not always hold true, unless the folid has fix faces: for a folid may be contained under parallel planes, and each of the oppofite planes will be indeed equal. But fome of them may not be parallelograms; as, if the eight angles of a cube were cut off by equal parallel planes, the remaining folid would confift of eight equal and oppofite triangles and eight equal and oppofite parallelograms or fquares.

PROP. XXV. THEOR.

If a folid parallelepipedon be cut by a plane, parallel to its oppofite planes: it will be as the bafe is to the bafe, fo is the folid to the folids.

For let the folid parallelepipedon ABCD be cut by the plane Y E parallel to the oppofite planes RA, DH: I fay as the base A E FQ is to the bafe EHCF, fo is the folid ABFY to the folid EGC D.

For produce AH both ways, and take any number of parts HM, MN each equal to EH, and any others AK, KL each equal to A E, and complete the parallelograms Lo, KQ, HW, MS, and the folids LP, KR, DM, MT. Then

because the right lines LK,

nother, the parallelograms

LO, KQ, AF, [by 38.1.] will

Q F C W S

be equal to one another; as al

fo the parallelograms KX, KB, AG; moreover the parallelograms L V, KP, AR are [by 24. II.] equal to one another; for they are oppofite parallelograms. By the fame reason the parallelograms E C, HW, Ms are equal to one another; as alfo the parallelograms HG, HI, IN, are equal. to one another: and fo are the parallelograms D H, MF,, NT: Therefore three planes of the folids L P, KR, AY are equal to three planes. But these three planes are equal to the three oppofite ones: Therefore [by 10. def. 11.] the three folids LP, KR, A Y will be equal to one another: By the fame reafon the three folids ED, DM, MT are equal to one another: Therefore the base LF is the fame mul tiple of the base AF, as the folid Ly is of the folid a Y. By the fame reason, the base NF is the fame multiple of the base HF, as the folid NY is of the folid HY. Again if the bafe LF be equal to the base N F, the folid L Y will be equal to the folid NY, and if L F exceeds the base N F, the folid Ly will exceed the folid NY; and if less, less. There

are