Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση
[ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

то

PROP. XXIII. P ROB.

O make a folid angle of three plain angles, any two of which are greater than the third; but these three angles muft, together, be less than four right angles.

a

It is required to make a folid angle of three plain angles ABC, DEF, GHK, any two of which are greater than the third; and all the angles together lefs than four right angles. Let the right lines AB, BC, DE, EF, GH, HK, be made equal to one another; and join AC, DF, GK. Then a triangle may be made of three right lines equal to AC, DF, GK; which let be LMN; make the fide LM equal to AC, MN to DF, and LN to GK ; defcribe the circle LMN about the triangle ; its center, X, is either within the triangle, upon one of the fides, or without the triangle. First, let it be within the triangle, and join LX, MX, NX; then, if AB be not greater than LX, it will be either equal or lefs. First, let it be equal; then AB, BC, are equal to LX, XM, and the bafe LM equal to AC; then the angle LXM is equal to the angle ABC 4. For the same reason, the angle MXN is equal to DEF, and NXL to GHK ; but the three angles LXM, MXN, NXL, are equal to four right angles; therefore ABC, DEF, GHK, are equal to four right angles; but they are lefs f; which is abfurd; therefore LX, XM, are not equal to AB, BC; and they are not greater. For, if poffible, let LX, XM, be greater than AB, BC, and cut off XO, XP, equal to AB, BC; join OP. Then, because AB is equal to BC, and XO to XP, the remainders LO, MP, will be equal; therefore OP is parallel to LM ; and the triangles LXM, OXP, are equiangular; therefore XO is to OP as XL is to LM: and, by altern. XO is to XL as OP is to LM. But LX is greater than XO; therefore LM is greater than OP. But LM is put equal to AC; therefore AC is greater than OP; therefore the angle ABC will be greater than the angle OXP. For the fame reason, DEF is greater than MXN, and GHK than NXL; but OXP, MXN, NXL, are equal to four right angles; therefore the angles ABC, DEF, GHK, are greater than four right angles, and likewife lefs f; which is impoffible : Therefore LX, XM, are not greater than AB, BC; but they are proved not equal; therefore they are lefs; therefore, on the point X, raise XR perpendicular to the plain of the circle LMN, and equal to the excefs by which the fquare of AB exceeds the fquare of LX; and join RL, RM, RN. Then, because RX is perpendicular to the plain LMN, it is at right angles to LX, MX, NX; therefore the fquares of LX, XR, are equal to the fquare

of

of LR". For the fame reafon, the fquares of RX, XM, are e- Book VI. qual to the fquare of RM; and the fquares of RX, XN, equal

to the fquare of RN; but LX, MX, NX, are equal, and RX n 47. I. common; therefore LR, RM, RN, are equal; but the fquare of AB is equal to the fquares of LX, XR; therefore LR is e- o conft. qual to AB. But BC, ED, EF, GH, HK, are each equal to AB; therefore RL, RM, RN, are each equal to AB or BC; and the base ML equal to AC; therefore the angle MRL is equal to the angle ABC; but MR, RN, are equal to DE, EF, and the base MN to DF, the angle MPN to DEF, and the angle LRN to GHK ; therefore the folid angle at R is contained by the three plain angles LRM, MRN, LRN, equal to the three plain angles ABC, DEF, GHK.

Now, let the center of the circle X be on one fide of the triangle, viz. MN; join XL; then AB is greater than LX. For, if not, it will be either equal or lefs. First, let AB be equal to LX; then MX, XL, are equal to AB, BC; that is, MX, XL, are equal to MN; but MN is equal to DF; therefore DE, EF, are equal to DF; which is abfurd P; much less can MX, XL, p 20. I. that is, MN, that is, DF, be greater than DE, EF; therefore AB is greater than LX; and, if XR is drawn perpendicular to the plain LMN, and equal to the excess by which the square of AB exceeds the fquare of LX, the figure can be constructed as before.

Laftly, let the center X of the circle be without the triangle LMN; join LX, MX, NX; then AB is greater than LX. If not, it is either equal or less. First, let it be equal; then the two fides AB, BC, are equal to the two fides MX, XL; and the base AC equal to ML; therefore the angle ABC is equal to the angle MXL d. For the fame reafon, GHK is equal to d 8. I. LXN; but the whole angle MXN is equal to the angles MXL, NXL; therefore MXN is equal to ABC, GHK; that is, DEF is equal to ABC, GHK; but ABC, GHK, is greater than DEF, and likewife equal; which is abfurd; therefore AB is 9 20. not equal to LX. Let AB be lefs than LX, and make OX, XP, equal to AB, BC; then the remainders OL, MP, will be equal; therefore OP is parallel to ML, and the triangles equi- 8 2. 6. angular; therefore XO is to OP as XL is to LM; by altern. as XO is to XL, fo is OP to LM; but XL is greater than XO; therefore LM is greater than OP; but LM is equal to AC; therefore AC is greater than OP; and the angle ABC greater than OXP*. Draw XV equal to XO or XP; and join OV; then the 25. I. angle GHK is greater than OXV. At the point X, with the right line LX, make the angle LXS equal to ABC; and the angle LXT to GHK'; and XS, XT, each equal to XO; and join 23. I. OS, OT, ST; then, because the two fides AB, BC, are equal

k

to

S 4. I.

t 34. I.

Book XI. to the two fides OX, XS, and the angle ABC to OXS; the base AC, that is, LM, will be equal to OSs. For the fame reafon, LN is equal to OT; and, fince the two fides ML, LN, are equal to the two fides OS, OT, and the angle MLN, or POV, greater than SOT, for it contains it, the base MN is greater than ST; but MN is equal to DF; therefore DF is greater than ST; therefore the angle DEF is greater than SXT; but the angle SXT is equal to the angles ABC, GHK; therefore the angle DEF is greater than ABC, GHK. and likewife lefs; which cannot be; therefore AB is not lefs than LX; but it has been proved not equal to it; therefore must be greater. Then, make XR equal to the excefs by which the fquare of AB exceeds the fquare of LX; and join RM, RN, RL; then, in the fame manner, it may be proved, that the folid angle R is the angle required. Wherefore, &c.

k 25. I.

[ocr errors]
[merged small][merged small][ocr errors][merged small][merged small]

PROP. XXIV. THE OR.

IF a folid be contained by fix parallel plains, the opposite plains thereof are equal parallelograms.

a

Let the folid CDGH be contained by the parallel plains AC, GF, BG, CE, FB, AE, the opposite plains thereof are equal parallelograms. For, becaufe the parallel plains BG, CE, are cut by the plain AC, their common fections AB, CD, are parallel ; and, because the parallcl plains BF, AE, are cut by the plain AC, their common fections AD, BC, are parallel; therefore AC is a parallelogram. In the fame manner, it is proved that GF is a parallelogram. Then, becaufe BH, AG, CF, DE, join the parallel lines AD, GE, BC, HF, they are equal to one another. For the fame reason, AB, HG, CD, EF, are equal to one another; therefore BG, CE, AC, GF, AE, BF, are parallelograms. Join AH, DF; then, becaufe AB, BH, are parallel to DC, CF, the angle ABH is equal to DCF; then, because AB, BH, are equal to DC, CF, and the angles ABH, DCF, equal, the bafes AH, DF, are equal d; but the parallelogram BG is double the triangle ABH, and CE double CDF; therefore the parallelograms BG, CE, are equal; in the fame manner, the parallelograms AC, GF, are proved equal; and AE equal to BF. Wherefore, &c.

PROP.

Book XI.

[ocr errors]

PRO P. XXV. THE OR.

a folid parallelopipedon be cut by a plain parallel to oppofite plains; then, as bafe is to bafe, fo is folid to folid.

Let the folid ABCD be cut by a plain YE, parallel to the oppofite plains RA, DH; then, as the base EFUA is to the bafe EHCF, fo is the folid ABFY to the folid EGCD. For, produce AH both ways, and make AK, KL, each equal to AE; and HM, MN, each equal to EH; and compleat the parallelograms LO, KU, HX, MS, and the folids LP, KR, HQ, MT; then, becaufe the right lines LK, KA, AE, are equal, the parallelograms LO, KU, AF, are equal; as also the pa- a 36. 1. rallelograms KV, KB, AG, and the parallelograms LW, KP, AR. For the fame reason, the parallelograms EC, HX, MS, are equal; as alfo, HG, HI, IN, and the parallelograms DH, MQ, NT. Then, becaufe LK, KA, AE, are equal; and likemife HM, MN, each equal to HE; LE is the fame multiple of AE that LF is of AF; and EN the fame multiple of EH that ES is of EC; and LG of AG; LR of AR; ET of HG; and EQ, of EY. Wherefore the three plains in the folid LP, and the three oppofite ones, which are equal to them, are equal to the three plains in the folid KR, or AY, and the three oppofite plains which are equal to them ; therefore the three folids b 24. LP, KR, AY, are equal, and the fame multiple of AY that e def. 1o. LF is of AF. For the fame reafon, the folids ED, HQ, MT, are equal; therefore ET is the fame multiple of ED that ES is of EC: Wherefore, if LF be equal to ES, the folid LY will be equal to the folid NY, if greater, greater, and, if lefs, lefs; Wherefore, as AF is to FH, fo is the folid AY to ED d. Where- & def. 5. 5, fore, &c.

A

PROP. XXVI. P ROB.

Ta given point, in a given right line, to make a folid angle
cqual to a folid angle given.

It is required, at a given point A, in a given right line AB, to make a folid angle equal to the folid angle contained by the plain angles EDC, EDF, FDC. In the right line DF affume any point F; from which draw FG perpendicular to the plain paffing through ED, DC, meeting the plain in the point

G;

a II.

h 23. I.

C 12.

d def. 3.

Book XI. G ; join DG; at the point A, with the right line AB, make the angles BAL, BAK, equal to the angles EDC, EDG; and make AK equal to DG; at the point K, in the plain BAL, raise a perpendicular HK: which make equal to GF; and join HA; then the folid angle at A, which is contained by the plain angles BAL, BAH, HAL, is equal to the folid angle. at D, contained by the plain angles EDC, EDF, FDC. For, take the right line AB, equal to DE; AL to DC; and join HB, KB, FE, GC, FC; then, because GF is perpendicular to the plain EDC, the angles FGD, FGE, FGC, are right angles; for the fame reafon, HKA, HKB, HKL, are right angles; and, because the two fides KA, AB, are equal to the two fides GD, DE, and contain equal angles, the bafes BK, EG, are equal; and, because BK, KH, are equal to EG, GF, each to each, and contain equal angles, the bafes HB, FE, are equal. Again, because AK, KH, are equal to DG, GF, each to each, and contain equal angles, the bale AH is equal to DF; but AB, AH, are equal to DE, DF, and the bafe BH equal to EF; therefore the angle BAH is equal to EDF f; but the angle BAL is equal to EDC, and a part BAK equal to EDG; therefore the remainders KAL, GDC, are equal, and the base KL to GC; and, because HK, KL, are equal to FG, GC, each to each, and the angle HKL equal to FGC, the bafe HL is equal to FC; but HA, AL, are equal to FD, DC, and the bafe HL equal to FC; the angle HAL is equal to FDC; therefore the plain angles BAL, BAH, HAL, containing the folid angle A, are equal to the plain angles EDC, EDF, FDC, containing the folid angle at D, each to each; therefore the folid angle at def. 10. A is made equal to the folid angle at D; which was to be done.

A. I.

[ocr errors]
[ocr errors]

то

PRO P. XXVII. PRO B.

O defcribe a parallelopipedon from a given right line, fimilar and alike fituate to a jolid parallelopipedon given.

It is required to describe, from the right line AB, a folid parallelopipedon, fimilar and alike fituate to the given folid parallelopipedon CD.

At the point A, in the given right line AB, make a folid angle A, contained by the plain angles BAH, HAK, KAB, equal to the folid angle at Ca, fo that the angle BAH be equal to ECF; BAK to ECG; and HAK to FCG; and make BA to AK

« ΠροηγούμενηΣυνέχεια »