c

Book XI. to the plane BAC, the plane HBK which paffes through HK is at right angles to the plane BAC; and AB is drawn in the plane b. 18. 11. BAC at right angles to the common fection BK of the two planes; .4.Def.11. therefore AB is perpendicular to the plane HBK, and makes right 3.Def.11. angles 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 reafon, DEM is a right angle, and is therefore equal to the angle ABH. and the angle HAB is equal to the angle MDE. therefore in the two triangles HAB, MDE there are two angles in one equal to two angles in the other, cach to each, and one fide equal to one fide, oppofite to one of the equal angles in each, viz. HA c. 26. 1. equal to DM; therefore the remaining fides are equal, each to each. wherefore AB is equal to DE. In the fame manner, if HC and MF be joined, it may be demonftrated that AC is equal to DF. therefore fince AB is equal to DE, BA and AC are equal

f. 4. 1.

G

to ED and DF; and the angle BAC is equal to the angle EDF; wherefore the base BC is equal to the bafe EF, and the remaining 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 fame reafon, 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 fide equal to one fide adjacent to the equal angles in each, viz. BC equal to EF; the other fides therefore are equal to the other fides; 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 bafe AK is equal to the bafe DN. and fince AH is equal to DM, the fquare of AH is equal to the

237

fquare of DM. but the fquares of AK, KH are equal to the fquare Book XI. of AH, becaufe AKH is a right angle. and the fquares of DN, NM are equal to the fquare of DM, for DNM is a right angle. g. 47. 1. wherefore the fquares of AK, KH are equal to the fquares of DN, NM; and of thofe the fquare of AK is equal to the fquare of DN. therefore the remaining fquare of KH is equal to the remaining fquare of NM; and the straight line KH to the straight line NM. and becaufe HA, AK are equal to MD, DN, each to each, and the b. fe HK to the bafe MN, as has been proved; therefore the angie HAK is equal to the angle MDN. Q. E. D.

COR. From this it is manifeft, that if from the vertices of two equal plane angles there be elevated two equal ftraight lines containing equal angles with the fides of the angles, each to each; the perpendiculars drawn from the extremities of the equal straight lines to the planes of the firft angles are equal to one another.

Another Demonftration of the Corollary.

Let the plane angles BAC, EDF be equal to one another, and let AH, DM be two equal ftraight 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.

h. 8. 1.

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 folid angles at Aand 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 AH coincides with DM, as was fhewn in Prop. B. of this Book. and becaufe AH is equal to DM, the point H coincides with the point M. wherefore HK which is perpendicular to the plane BAC coincides with MN i. 13. 11. 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.

See N.

IF

PROP. XXXVI. THEOR.

F three ftraight lines be proportionals, the folid parallelepiped defcribed from all three as its fides, is equal to the equilateral parallelepiped defcribed from the mean proportional, one of the folid angies of which is contained by three plane angles equal, each to each, to the three plane angles containing one of the folid angles of the other figure.

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

Take a folid angle D contained by three plane angles EDF, FDG, GDE; and make each of the straight lines ED, DF, DG equal to B, and complete the folid parallelepiped DH. make LK a. 26. 11. equal to A, and at the point K in the ftraight line LK make a folid angle contained by the three plane angles LKM, MKN, NKL equal to the angles EDF, FDG, GDE, each to each; and make

H

b. 14. 6.

KN equal to B, and KM equal to C; and complete the folid parallelepiped KO. and becaufe, as A is to B, fo is B to C, and that A is equal to LK, and B to each of the ftraight lines DE, DF, and C to KM; therefore LK is to ED, as DF to KM; that is, the fides about the equal angles are reciprocally proportional; therefore the parallelogram LM is equal 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

from the points G, N, to the planes EDF, LKM are equal to Book XI. one another. therefore the folids KO, DH are of the fame altitude; and they are upon equal bafes LM, EF, and therefore they are c. Cor. 35equal to one another. but the folid KO is defcribed from the three ftraight lines A, B, C, and the folid DH from the straight line B. If therefore three ftraight lines, &c. Q. E. D.

IF

II.

d. 31. 11.

F four ftraight lines be proportionals, the fimilar folid See N. parallelepipeds fimilarly defcribed from them fhall alfo be proportionals. and if the fimilar parallelepipeds fimilarly defcribed from four straight lines be proportionals, the ftraight lines fhall be proportionals.

Let the four straight lines AB, CD, EF, GH be proportionals, viz. as AB to CD, fo EF to GH; and let the fimilar parallelepipeds AK, CL, EM, GN be fimilarly defcribed from them. AK is to CL, as EM to GN.

Make AB, CD, O, P continual proportionals, as alfo EF, a. 11. 6. GH, Q, R. and because as AB is to CD, fo EF to GH; CD

is to O, as GH to Q, and O to P, as Q to R; therefore, ex b. 11. 5.

aequali, AB is to P, as EF to R. but as AB to P, fo is the c. 22. 5. folid AK to the folid CL; and as EF to R, fo is the folid EMd. Cor. 33. to the folid GN. therefore as the folid AK to the folid CL, fo is the folid EM to the folid GN.

JI.

But let the folid AK be to the folid CL, as the folid EM to the folid GN. the ftraight line AB is to CD, as EF to GH.

e

Take AB to CD, as EF to ST, and from ST defcribe a folid parallelepiped SV fimilar and fimilarly fituated to either of the folids EM, GN. and because AB is to CD, as EF to ST, and that from AB, CD the folid parallelepipeds AK, CL are fimilarly defcribed; and in like manner the folids EM, SV from the ftraight lines EF,

鸭鸭鸭

E

T

G HQ R

ST; therefore AK is to CL, as EM to SV. but, by the hypothefis, AK is to CL, as EM to GN. therefore GN is equal f to SV. but it is likewife fimilar and fimilarly fituated to SV; therefore the planes which contain the folids GN, SV are fimilar and equal, and their homologous fides GH, ST equal to one another. and becaufe as AB to CD, fo 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.

PROP. XXXVIII. THEOR.

IF a plane be perpendicular to another plane, and a ftraight line be drawn from a point in one of the planes perpendicular to the other plane, this ftraight "line fhall fall on the common fection of the planes.

ઃઃ

"Let the plane CD be perpendicular to the plane AB, and let "AD be their common fection; if any point E be taken in the "plane CD, the perpendicular drawn from E to the plane AB "fhall fall on AD.