PROP. A. THEOR.

If each of two solid angles be contained by three 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 (9. def. 11.) of the plane CAD to the plane CAE. In BF take BM equal to AK, and from the point

A

B

M draw, in the planes FBG, FBH, the straight lines MG, MN at right angles to BF; therefore the angle GMN is the inclination (6. def. 11.) 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 equal (26. 1.) to MG, and AD to BG: for the same reason, in 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 (4. 1.) to the base NG. Lastly, in the triangles 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 (8. 1.) to the angle GMN: but the angle DKL is the inclination of the plane CAD to the plane CAE, and the angle GMN is the inclination of the

* See Note.

plane FBG to the plane FBH, which planes have therefore the same inclination (7. def. 11.) to one another: and in the same manner it may be demonstrated, that the other planes in which the equal angles are, have the same inclination to one another. Therefore, if two solid angles, &c. Q. E. D.

PROP. B. THEOR.

Ir two solid angles be contained, each by three plane angles which are equal to one another, each to each, and alike situated; these solid angles are equal to one another.*

Let there be two solid angles at A and B, of which the solid angle at A is contained by the three plane angles CAD, CAE, EAD; and that at B, by the three plane angles FBG, FBH, HBG; of which CAD is equal to FBG; CAE to FBH; and EAD to HBG: the solid angle at A is equal to the solid angle

at B.

A

B

Let the solid angle at A be applied to the solid angle at B; and, first, the plane angle CAD being applied to the plane angle FBG, so as the point A may coincide with the point B, and the straight line AC with BF; then AD coincides with BG, because the angle CAD is equal to the angle FBG; and because the inclination of the planeCAE to the planeCAD is equal (A. 11.) to the inclination of the plane FBH to the plane FBG, the plane CAE coincides with the plane FBII, because the planes CAD,

ΛΑ

C

E

F

G

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 are equal (A. 8. 1.) to one another. Q. E. D.

See Note.

PROP. C. THEOR.

SOLID figures 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; 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 (B. 11.) to the solid angle at K: in the same manner, 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 figure AC being ap

plied to the plane figure KM: the straight line AB coinciding with KL, D the figure AC must coincide with the figure KM, because

H

R

Q

A

G

В

K

P

M

L

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 (B. 11.) 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 0, 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

• See Note.

figure LQ, and the straight line CG with MQ, and the point G 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.

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.

B

H

Because the parallel planes BG, CE are cut by the plane AC, their common sections AB, CD are parallel (16. 11.). Again, because the two parallel planes BF, AE are cut by the plane AC, their common section AD, BC are parallel (16. 11.); and AB is parallel to CD; therefore AC is a parallelogram. In like manner, it may be proved that each of the figures CE, FG, GB, BF, AE is a parallelogram: Join AH, DF; and because AB is parallel to DC, and BH to CF; the two straight lines AB, BH, which meet one another, are parallel to DC and CF which meet one another, and are not in the same plane D with the other two: wherefore they contain equal angles (10. 11.); the angle

A

E

G

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 (4. 1.) to the base DF, and the triangle ABH to the triangle DCF: and the parallelogram BG is double (34. 1.) 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

* See Note.

is equal and similar to the parallelogram GF, and the parallelogram AE to BF. Therefore, if a solid, &c. Q. E. D.

PROP. XXV. THEOR.

Ir a solid parallelopiped be cut by a plane parallel to two 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 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, the parallelograms

LO, KY, AF are equal (36. 1.): and likewise the parallelograms KX, BK, AG, (36. 1.); as also (24. 11.) the parallelograms LZ, KP, AR, because they are opposite planes: for the same reason the parallelograms EC, HQ, MS are equal (36. 1.); and the parallelograms HG, HI, IN, as also (24. 11.) 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 (24. 11.) 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 (C. 11.) to one another: for the same reason, the three solids ED, HU, MT are equal to one another therefore what multiple soever the base LF

* See Note.