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Book XI.

PROP. A.

THEOR.

I

F 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 of the plane a.6.Def. pl. CAD to the plane CAE. in BF take BM equal to AK, and from the point M draw, in the planes FBG, FBH, the straight lines MG, MN at right an. gles to BF; therefore

A

B the angle GMN is the inclination a of the plane FBG to the plane FBH.

K

M

N join LD, NG; and beCause in the triangles C

D

E KAD, MBG, the angles E

H 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 to MG, and AD to BG. for the same reason, in the b. 26. E. 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 C. 4. 7. 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 d to the angle GMN. but the d. 8. 1. angle DKL is the inclination of the plane CAD to the plane CAE,

AG

Book XI. and the angle GMN is the inclination of the plane FBG to the

plane FBH, which planes have therefore the same inclination to e.7.Def.nn. 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.

See N.

F 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 folid 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.

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

A B angle FBG. and because the in

clination of the plane CAE to the 3. A. 11. plane CAD is equal to the inclination of the plane FBH to the

E plane FBG, the plane CAE coin

F

H

D Ġ cides 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 folid angle A coincides with the 6. 8. Ax. I. folid angle B, and consequently they are equal to one another,

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Q. E. D.

Book XI.

PROP. C.

THEOR.

OLID figures contained by the fame number of equal see N.

S

of their folid 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 num. ber 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
angies LKN, LKO, OKN which contain the solid angle at K, each
to each; therefore the solid angle at A is equal a to the folid angle a. Bo 11.
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 applied to the
plane figure KM; the
straight line AB coin-H G R
ciding with KL, the

F O
figure AC muft co-
incide with the figured

C N! M
KM, because they
are equal and similar.

А B K L
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
folid angle at A coincides a with 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 ano-
ther. 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 folid angle at B is equal to the solid angle at L, it
may be proved in the same manner, that the figure LG coincides

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Book XI. with the figure LQ, and the straight line CG with MQ, and

the point G with the point Q. since therefore all the planes and
fides of the folid figure AG coincide with the planes and sides of
the solid figure KQ, AG is equal and fimilar to KQ. and in the
same manner, any other solid figures whatever contained by the
same number of equal and fimilar 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.

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See N.

F a solid be contained by fix 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 plane #. 16. 11. AC, their common sections AB, CD are parallel *, again, because

the two parallel planes BF, AE are cut by the plane AÇ, their
conimon sections AD, BC are parallela. and AB is parallel to CD;
therefore AC is a parallelogram. in

B H
like manner, it may be proved that
each of the figures CE, FG, GB, BF, A

G
AE is a parallelogram. join AH, DF;
and because AB is parallel to DC, and
BH to CF; the two straight lines AB,

F
EH, which meet one another, are pa-
rallel to DC and CF which meet one D

another and are not in the same plane
D. 10. 11. with the other two; wherefore they contain 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 A BH equal to the angle C. 4. I. DCF, therefore the base AH is equal « to the bafe. DF, and the

triangle ABH to the triangle DCF. and the parallelogram BG is 4. 34. 1.

double d of the triangle A BH, 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 paralļelogram AC is equal and fimilar to the pas

rallelogram GF, and the parallelogram AE to BF. Therefore if a Book XI. folid, &c. Q. E. D.

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F a solid parallelepiped be cut by a plane parallel to See N.

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 folid is to the other.

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Let the solid parallelepiped 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 , a. 36. I.

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0 Y F

C Q

S and likewise the parallelograms KX, KB, AG •; as also the pa- bo 24. 11. rallelograms LZ, KP, AR, because they are opposite planes. for the same reason, the parallelograms EC, HQ , MS are equal ~; and the parallelograms HG, HI, IN, as also b HD, MU, NT. therefore three planes of the folid LP, are equal and similar to three planes of the folid KR, as also to three planes of the folid AV. but the three planes opposite to these three are equal and similar 6 to them in the several folids, 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, the c. C. II. three solids ED, HU, MT are equal to one another. therefore-what

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