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Book XI. to the plane BAC, the plane HBK which passes through HK is
meets it in that plane; therefore ABH is a right angle. for the
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
to ED and DF; and the angle BAC is equal to the angle EDF;
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.
PRO P. XXXVI. THEO R.
rallelepiped described from all three as its sides, is
Let A, B, C be three proportionals, viz. A to B, as B to C.
Take a solid angle D contained by three plane ancies EDF,
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
KN equal to B, and KM equal to C; and complete the folid pa-
b. 14. 6.
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.
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.
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,
E к G H Q R
PRO P. XXXVIII. THEOR.
straight line be drawn from a point in one of the
“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.