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Book XI. to the plane BAC, the plane HBK which passes through HK is
Wat right angles b to the plane BAC; and AB is drawn in the plane
B. 18. 11. BAC at right angles to the common section BK of the two planes;
c.4. Def.11. therefore AB is perpendicular to the plane HBK, and makes right
d. 3. Def. 11. angles d with every straight line meeting it in that plane. but BH

meets it in that plane; therefore ABH is a right angle. for the
fame reason, DEM is a right angle, and is therefore equal to the
angle ABH. and the angle HAB is equal to the angle MDE. there-
fore in the two triangles HAB, MDE thcre are two angles in one
equal to two angles in the other, cach to each, and one side equal

to one side, opposite to one of the equal angles in each, viz. HA 6. 26. 1. equal to DM ; therefore the remaining sides are equal“, each to

each. wherefore AB is equal to DE. In the fame manner, if
HC and MF be joined, it inay be demonstrated that AC is equal
to DF. therefore fince AB is equal to DE, BA and AC are equal

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to ED and DF; and the angle BAC is equal to the angle EDF;
wherefore the base BC is equal é to the base EF, and the remain-
ing angles to the remaining angles. the angle ABC is, therefore
equal to the angle DEF. and the right angle ABK is equal to the
right angle DEN, whence the remaining angle CBK is equal to
the remaining angle FEN. for the same reason, the angle BCK is
equal to the angle EFN. therefore in the two triangles BCK, EFN
there are two angles in one equal to two angles in the other, each
to each, and one side equal to one side adjacent to the equal angles
in each, viz. BC equal to EF; the other sides therefore are equal
to the other sides; BK then is equal to EN. and AB is equal to
DE; wherefore AB, BK are equal to DE, EN; and they contain
right angles; wherefore the base AK is equal to the base DN.
and since AH is equal to DM, the square of AH is equal to the

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square of DM. but the squares of AK, KH are equal to the square Book XI. 8 of AH, because AKH is a right angle. and the squares of DN, NM are equal to the square of DM, för DNM is a right angle. 3. 47. 1. wherefore the squares of AK, KH are equal to the squares of DN, NM; and of those the square of AK is equal to the square of DN. therefore the remaining square of KH is equal to the remaining square of NM; and the straight line KH to the straight line NM. and because HA, AK are equal to MD, DN, each to each, an i the b. fe HK to the base MN, as has been proved; therefore the angie HAK is equal h to the angle MDN. Q. E. D.

h. 8. 11

Cor. From this it is manifest, that if from the vertices of two equal plane angles there be elevated two equal straight lines containing equal angles with the sides of the angles, each to each; the perpendiculars drawn from the extremities of the equal straight lines to the planes of the first angles are equal to one another.

Another Demonstration of the Corollary.

Let the plane angles BAC, EDF be equal to one another, and let AH, DM be two equal straight lines above the planes of the angles, containing equal angles with BA, AC, ED, DF, each to each, viz. the angle HAB equal to MDE, and HAC equal to the angle MDF; and from H, M let HK, MN be perpendiculars to the planes BAC, EDF; HK is equal to MN.

Because the folid angle at A is. contained by the three plane angles BAC, BAH, HAC, which are, each to each, equal to the three plane angles EDF, EDM, MDF containing the folid angle at D; the solid angles at A and D are equal, and therefore coincide with one another ; to wit, if the plane angle ABC be applied to the plane angle EDF, the straight line Al coincides with DM, as was thewn in Prop. B. of this Book. and because All is equal to DM, the point H coincides with the point M. wherefore HK which is perpendicular to the plane BAC coincides with i MN i. 13. IT. which is perpendicular to the plane EDF, because these planes coincide with one another, therefore HK is equal to MN. Q. E. D.

Book XI.

PRO P. XXXVI. THEO R.

See N.

IF
F three straight lines be proportionals, the solid pa-

rallelepiped described from all three as its sides, is
equal to the equilateral parallelepiped described from
the mean proportional, one of the folid angics of which
is contained by three plane angles equal, each to each,
to the three plane angles containing one of the solid
angles of the other figure.

Let A, B, C be three proportionals, viz. A to B, as B to C.
The solid described from A, B, C is equal to the equilateral fold
described from B, equiangular to the other.

Take a solid angle D contained by three plane ancies EDF,
FDG, GDE ; and make each of the straight lines ED, DF, DG

equal to B, and complete the solid parallelepiped DH. make LK 2. 26. 11. equal to A, and at the point K in the straight line LK make

solid angle contained by the threc plane angles LKM, MKN, NKL
equal to the angles EDF, FDG, GDE, each to each ; and make

H

a

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KN equal to B, and KM equal to C; and complete the folid pa-
rallelepiped KO. and because, as A is to B, so is B to C, and that
A is equal to LK, and B to each of the straight lines DE, DF, and
C to KM; therefore LK is to ED, as DP to KM; that is, the
fides about the equal angles are reciprocally proportional; there-
fore the parallelogram LM is equal o to EF. and because EDF,
LKM are two equal plane angles, and the two equal straight lines
DG, KN are drawn from their vertices above their planes, and
contain equal angles with their fides; therefore the perpendiculars

b. 14. 6.

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from the points G, N, to the plancs EDF, LKM are equal to Book XI. one another. therefore the fulids KO, DII are of the fame altitude; and they are upon equal bases LM, EF, and therefore they are c. Cor. 35. equal to one another. but the fulid KO is described froin the

d. 31. II. three straight lines A, B, C, and the solid DH from the straight line B. If therefore thiee straight lines, &c. Q. E. D.

PRO P. XXXVII.

THEOR. If four straight lines he proportionals, the similar solid sce N.

parallelepipeds similarly described from them shall also be proportionals. and if the similar parallelepipeds similarly described from four straight lines be proportionals, the straight lines fhall be proportionals.

Let the four straight lines AB, CD, EF, GH be proportionals, viz. as AB to C!, fo EF to GH; and let the similar parallelepipeds AK, CL, EM, GN be similarly described from them. AK is to CI., as EM to GN.

Make AB, CO, O, P continu il proportionals, as also EF, a. 11. 6. GH, Q, R. and because as AB is to CD, fo EF to GH; CD is to , as GH to Q, and O to P, as to R; therefore, ex b. 11. s.

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S T
E

к G 1 Q R aequali“, AB is to P, as EF to R. but as AB to P, so d is the c. 22. s. folid AK to the solid CL; and as EF to R, fod is the solid EM d. Cor. 3 3. to the folid GN. therefore b as the solid AK to the folid CL, so is the folid EM to the solid GN.

II.

e. 27. II.

Book XI. But let the solid AK be to the folid CL, as the solid EM to the w folid GN. the straight line AB is to CD, as EF to GH.

Take AB to CD, as EF to ST, and from ST describe e a folid parallelepiped SV similar and similarly situated to either of the folids EM, GN. and because AB is to CD, as EF to ST, and that fron AB, CD the folid parallelepipeds AK, CL are similarly described; and in like manner the folids EM, SV from the straight lines EF,

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1.9. 5.

E к G H Q R
ST; therefore AK is to CL, as EM to SV. but, by the hypothesis,
AK is to CL, as EM to GN. therefore GN is equal to SV. but
it is likewise similar and similarly situated to SV; therefore the
planes which contain the solids GN, SV are similar and equal, and
their homologous sides GH, ST equal to one another. and be-
cause as AB to CD, so EF to ST, and that ST is equal to GH;
AB is to CD as EF to GH. Therefore, if four straight lines, &c.
Q. E. D.

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See N.

IF

PRO P. XXXVIII. THEOR.
F a plane be perpendicular to another plane, and a a

straight line be drawn from a point in one of the
planes perpendicular to the other plane, this straight
“ line shall fall on the common section of the planes.

“Let the plane CD be perpendicular to the plane AB, and let “ AD be their common section ; if any point E be taken in the “ plane CD, the perpendicular drawn from E to the plane AB « shall fall on AD.

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