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PROP. A. THEOR.
If each of two solid angles be contained by three See N. plane angles equal to one another, each to each ; the planes in which the equal angles are, have the same inclination to one another.
Let there be two solid angles at the points A, B; and let the angle at A be contained by the three plane angles CAD, CAE, EAD; and the angle at B by the three plane angles FBG, FBH, HBG; of which the angle CAD is equal to the angle FBG, and CAE to FBH, and EAD to HBG: The planes in which the equal angles are, have the same inclination to one another,
In the straight line AC take any point K, and in the plane CAD from K draw the straight line KD at right angles to AC, and in the plane CAE the straight line KL at right angles to the same AC: Therefore the angle DKL is the inclination a of
1 6 Def. 11. the plane CAD to the plane CAE: In BF take BM
K equal to AK, and
MA from the point M C draw, in the planes FBG, FHB, the
H Η straight lines MG, MN at right angles to BF; therefore the angle GMN is the inclination a of the plane FBG to the plane FBH: Join LD, NG; and because in the triangles KAD, MBG, the angles KAD, MBG are equal, as also the right angles AKD, BMG, and that the sides AK, BM, adjacent to the equal angles, are equal to one another; therefore KD is equalb to MG, and AD to BG: For the same reason, in ) 26. 1. the triangles KAL, MBN, KL is equal to MN, and AL to BN: And in the triangles LAD, NBG, LA, AD are equal to NB, BG, and they contain equal angles : therefore the base LD is equal to the base NG. Lastly, in the triangles • 4. 1. KLD, MNG, the sides DK, KL are equal to GM, MN, and the base LD to the base NG; therefore the angle DKL is equal tod the angle GMN: But the angle DKL 8. 1. is the inclination of the plane CAD to the plane CAE, and the angle GMN is the inclination of the plane FBG to the
Boox XI. plane FBH, which planes have therefore the same inclinationa to one another; And in the same manner it may
be • 7 Def. 11. demonstrated, that the other planes in which the equal an
gles are, have the same inclination to one another. There-
See N. If two solid angles be contained, each by three
plane angles which are equal to one another, each
Let there be two solid angles at A and B, of which the
*Let the solid angle at A be applied to the solid angle at
nation of the plane CAE to the • A. 11. plane CAD is equal to the inclination of the plane FBH to
С the plane FBG, the plane CAE
G coincides with the plane FBH, because the planes CAD, FBG coincide with one another : And because the straight lines AC, BF coincide, and that the angle CAE is equal to the angle FBH; therefore AE coincides with BH, and' AD coincides with BG; wherefore the plane EAD coincides with the plane HBG: Therefore the solid angle A
coincides with the solid angle B, and consequently they 18A. 1. are equal b to one another, Q.E.D.
PROP. C. THEOR.
SOLID figures contained by the same number of See N. equal and similar planes alike situated, and having none of their solid angles contained by more than three plane angles, are equal and similar to one another.
Let AG, KQ be two solid 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 similar and equal to PR: The solid figure AG is equal and similar to the solid figure KQ.
Because the solid 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 a to the solid angle at K: In the .B. 11. same manner, the other solid angles of the figures are equal to one another. If, then, the solid figure ÅG be applied to the solid fiH
R gure KQ, first the plane figure E
P AC being ap
IC plied to
M plane figure KM; the straight A
K line AB coinciding with KL, the figure AC must coincide with the figure KM, because they are equal and similar: Therefore the straight lines AD, DC, CB coincide with KN, NM, ML, each with each: and the points, A, D, C, B, with the points K, N, M, L: And the solid angle at A coincides with a the solid angle at K; wherefore the plane AF coincides with the plane KP, and the figure AF with the figure KP, because they are equal and similar to one another : Therefore the straight lines AE, EF, FB, coincide with KO, OP, PL; and the points E, F, with the points O, P. In the same manner, the figure AH coincides with the figure KR, and the straight line DH with NR, and the point H with the point R: And because the solid angle at B is equal to the solid angle at L, it may be proved, in the same manner, that the figure BG coincides with the figure
Book XI. LQ, and the straight line CG with MQ, and the point G r with the point Q : Since, therefore, all the planes and sides
of the solid figure AG coincide with the planes and sides of the solid 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 situated, and having none of their solid angles contained by more than three plane angles, may be proved to be equal and similar to one another. Q.E.D.
PROP. XXIV. THEOR.
Şee N. IF a solid be contained by six planes, two and two
of which are parallel; the opposite planes are similar and equal parallelograms.
Let the solid 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. 11. plane AC, their common sections AB, CD, are parallela.
Again, because the two parallel planes BF, AE are cut by
same plane with the other two: wherefore they contain • 10. 11. 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 4. 1. base AH is equal to the base DF, and the triangle ABH & 34. 1. to the triangle DCF: And the parallelogram BG is doubled
of the triangle ABH, and the parallelogram CE double of the triangle DCF; therefore the parallelogram BG is equal and similar to the parallelogram CE. In the same manner it
may be proved, that the parallelogram AC is equal and
similar to the parallelogram GF, and the parallelogram Boor XI. AE to BF. Therefore, if a solid, &c. Q. E. D.
PROP. XXV. THEOR.
If a solid parallelopiped be cut by a plane parallel See N.
Let the solid parallelopiped ABCD be cut by the plane EV, which is parallel to the opposite planes AR, HD, and divides the whole into the two solids ABFV, EGCD; as the base AEFY of the first is to the base EHCF of the other, so is the solid 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, X
C Q S the parallelograms LO, KY, AF, are equal a: And likewise - 36. 1. the parallelograms KX, KB, AG“, as also b the parallelo- h 24. 11. grams LZ, KP, AR, because they are opposite planes: For the same reason, the parallelograms EC, HQ, MS, are equala; and the parallelograms HG, HI, IN, as also b 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 similarb to them in the several solids, and none of their solid 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. 11. the three solids ED, HU, MT, are equal to one another : Therefore what multiple soever the base LF is of the base