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PRO P. C.
OLID figures contained by the same number of e. See
none of their solid angles contained by more than three plane angles ; are equal and similar to one another.
Let AG, KQ be two folid figures contained by the same number of similar and equal planes, alike situated, viz. let the plane AC be similar and equal to the plane KM; the plane AF to KP; BG to LQ; GD to QN; DE to NO, and lastly, FH fimilar and equal to PR: The solid figure AG is equal and fimilar to the solid figure KQ.
Because the folid angle at A is contained by the three plane angles BAD, BAE, EAD, which, by the hypothesis, are equal to the plane angles LKN, LKO, OKN, which contain the solid angle at K, each to each ; therefore the solid angle at A is equal to the solid angle at K: In the same manner, a B. isi the other solid angles of the figures are equal to one another. If then the solid figure AG be applied to the solid figure KQ, first, the plane fi
H gure AC being
G R Q applied to the
o plane figure KM;
Book XI. the figure LQ, and the straight line CG with MQ, and the mpoint G with the point Q: Since therefore all the planes
and sides of the folid figure AG coincide with the planes and sides of the folid figure KQ, AG is equal and similar to KQ: And, in the same manner, any other solid figures whatever contained by the same number of equal and similar planes, alike fituated, and having none of their folid angles contained by more than three plane angles, may be proved to be equal and similar to one another. Q. E. D.
which are parallel ; the opposite planes are similar and equal parallelograms.
Let the folid CDGH be contained by the parallel planes AC, GF; BG, CE; FB, AE: Its opposite planes are similar and equal parallelograms.
Because the two parallel planes, BG, CE are cut by the @ 16. II.
plane AC, their common sections AB, CD are parallel. A. gain, because the two parallel planes BF, AE are cut by the plane AC, their common sections AD, BC are parallel * : And AB is parallel to CD; therefore AC is a parallelogram. In like manner, it may be proved that each
F and CF which meet one another, and are not in the same plane with
D. the other two; wherefore they conb 10.11. tain equal angles b; the angle ABH is therefore equal to the
angle DCF: And because AB, BH are equal to DC, CF, and the angle ABH equal to the angle DCF; therefore the base
AH is equal to the base DF, and the triangle ABH to the tri& 34. 1. angle DCF: And the parallelogram BG is doubled of the tri
angle ABH, and the parallelogram CE double of the triangle DCF; therefore the parallelogram BG is equal and fimilar to the parallelogram CE. In the same manner it may be proved, that the parallelogram AC is equal and fi
milar to the parallelogram GF, and the parallelogram AE to Book XI. BF. Therefore, if a solid, &c. Q. L. D.
of its opposite planes; it divides the whole into two solids, the base of one of which shall be to the base of the other, as the one solid is to the other.
Let the folid parallelepiped ABCD be cut by the plane EV, which is parallel to the opposite planes AR, HD, and divides the whole into the two folids ABFV, EGCD; as the base AEFY of the first is to the base EHCF of the other, so is the folid ABFV to the solid EGCD.
Produce AH both ways, and take any number of straight lines HM, MN each equal to EH, and any number AK, KL each equal to EA, and complete the parallelograms LO, KY, HQ, MS, and the solids LP, KR, HU, MT: Then, because the straight lines LK, KA, AE are all equal, the parallelograms
LO, KY, AF are equala: And likewise the parallelograms KX, a 36. 1. KB, AG*; as also b the parallelograms LZ, KP, AR, because b 24.11. they are opposite planes: For the same reason, the parallelograms EC, HO, MS are equal” ; and the parallelograms HG, HI, IN, as alio 6 HD, MU, NT: Therefore three planes of the solid LP, are equal and similar to three planes of the solid KR, as also to three planes of the solid AV: But the three planes opposite to these three are equal and similar to them in the several solids, and none of their folid angles are contained by more than three plane angles: Therefore the three solids LP, KR, AV are equal to one another : for the same reason, C. II. the three solids ED, HU, MT are equal to one another: ThereP
Book XI. fore what multiple foever the base LF is of the base AF, the
fame multiple is the folid LV of the solid AV: For the same reason, whatever multiple the base NF is of the base HF, the
same multiple is the solid NV of the solid ED: And if the base & C. ss.
LF be equal to the base NF, the solid LV is equal to the folid NV, and if the base LF be greater than the base NF, the folid LV is greater than the folid NV, and if less, lefs : Since then there are four magnitudes, viz. the two bases AF, FH,
Y F C Q. S and the two folids AV, ED, and of the base AF and folid AV, the base LF and solid LV are any equimultiples whatever; and of the base FH and solid ED, the base FN and solid NV are apy equimultiples whatever ; and it has been proved, that if the base LF is greater than the base FN, the folid LV is greater
than the solid NV ; and if equal, equal; and if lefs, less. d s. def. s. Therefored as the base AF is to the base FH, so is the solid AV
to the folid ED. Wherefore, if a solid, &c. Q. E. D.
See N. T a given point in a given straight line, to make a
solid angle equal to a given solid angle contained by three plane angles.
Let AB be a given straight line, A a given point in it, and D a given solid angie contained by the three plane angles EDC, EDF,FDC: It is required to make at the point À in the ftraight line AB a solid angle equal to the solid angle D.
In the straight line DF take any point F, from which drar : 11. 11. FG perpendicular to the plane EDC, meeting that plane in
G; join DG, and at the point A in the straight line AB 1 23. 1. make b the angle BAL equal to the angle EDC, and in the
plane BAL make the angle BAK equal to the angle EDG; then make AK equal to DG, and from the point KerectKH
at right angles to the plane BAL; and make KH equal to Book XI: GF, and join AH: Then the solid angle at A, which is contain m ed by the three plane angles BAL, BAH, HAL, is equal to the solid angle at D contained by the three plane angles EDC, EDF, FDC.
Take the equal straight lines AB, DE, and join HB, KB, FE, GE: And because FG is perpendicular to the plane EDC, it makes right angles d with every straight line meeting it in d 3. def. 11. chat plane: Therefore each of the angles FGD, FGE is a right angle: For the same reason, HKA, HKB are right angles : And because KA, AB are equal to GD, DE, each to each, and contain equal angles, therefore the base BK is equal to the c 4.1. base EG : And KH is equal to GF, and HKB, FGE, are right angles, therefore HB is equal to FE: Again, because AK, KH are equal to DG, GF, and contain right angles, the base AH is equal to the base DF; and AB is equal to DE; therefore HA, AB are equal to FD, DE, and the base HB is equal to the base FE, therefore the angle
f 8.s. BAH is equal f to the angle EDF: For the same reason, the angle HAL is equal
B to the angle FDC.
G and KL, HL, GC, FC be joined, since the whole angle BAL is equal to the whole EDC, and the parts of them BAK, EDG are, by the construction, equal; therefore the remaining angle KAL is equal to the remaining angle GDC: And because KA, AL are equal to GD, DC, and contain equal angles, the base KL is equal to the base GC: and KH is equal to GF, so that LK, KH are equal to CG, GF, and they contain right angles; therefore the base HL is equal to the base FC: Again, because HA, AL are equal to FD, DC, and the base HL to the base FC, the angle HAL is equalf to the angle FDC: Therefore, because the three plane angles BAL,BAH, HAL, which contain the solid angle at A, are equal to the three plane angles EDC, EDF, FDC, which contain the solid angle at D, each to each, and are situated in the same order, the solid angle at A is equal 6 to the folid angle at D. Therefore, at a given point in g B. II. a given straight line, a solid angle has been made equal to a given solid angle contained by three plane angles. Which was to be done.