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Book XI. Because HK is perpendicular to the plane BAC, the plane WHBK which passes through HK is at right angles to the plane b 18. 11, BAC; and AB is drawn in the plane BAC at right angles to
the common section BK of the two planes ; therefore AB is 44. def. 11, perpendicular to the plane HBK, and makes right angles
& d 3. def. Iv. with every straight line meeting it in that plane : But BH meets
it in that plane; therefore ABH is a right angle : For the same reason, 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, each to each, and one side equal to one fide, opposite to one of the equal angles
in each, viz. HA equal to DM; therefore the remaining Gdes C 26. 2. are equal, each to each: Wherefore AB is equal to DĒ. In
the same manner, if HC and MF be joined, it may be demonstrated that AC is equal to DF: Therefore, fince AB is equal to DE, BA and AC are equal to ED and DF; and the angle
BAC is equal to the angle EDF; wherefore the base BC is equalf to the base 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 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 que equal to one side adjacent to the equal angles in each, viz. BC equal to EF; the other fides, 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 dnce AH is equal to DM, the square of AH is equal to the square of DM: But the Book XI. squares of AK, KH are equal to the square 8 of AH, because w AKH is a right angle: And the squares of DN, NM are equal 8 47. I. to the square of DM, for DNM is a right angle: 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 remain- & ing 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, and the base HK to the base MN, as has been proved, therefore the angle HAK is equal to the angle MDN. h 8. 1. E. D.
f 4. I.
Cor. From this it is manifeft, 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 plaries 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 angies EDF, EDM, MDF containing the solid angle at D; the solid angles at A and D are equal: And therefore coincide with one another ; to wit, if the plane angle BAC be applied to the plane angle EDF, the straight line AH coincides with DM, as was shewn in prop. B of this book : And because AH 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 which is perpendicular to the plane i 13. if. EDF, because these planes coincide with one another : Therefore HK is equal to MN. 0. E. D.
PRO P. XXXVI.
F three straight lines be proportionals, the solid paral
lelepiped described from all three as its fides, is equal to the equilateral parallelepiped described from the mean proportional, one of the solid angles 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 described from A, B, C is equal to the equila. teral solid described from B, equiangular to the other.
Take a solid 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 solid parallelepiped DH :
Make LK equal to A, and at the point K’in the straight line a 26. 11. LK make a solid angle contained by the three plane angles
LKM, MKN, NKL equal to the angles EDF, FDG, GDE,
each to each ; and make KN equal to B, and KM, equal to C; and complete the solid parallelepiped KO: And because, as A is to B, co 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 DF to KM; that is, the sides about the
equal angles are reciprocally proportional; therefore the pab 14. 6. rallelogram 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 perpendicu. lars from the points G, N, to the planes EDF, LKM are e
c Cor. 35
qual to one another : Therefore the solids KO, DH are of Book XI.
PRO P. XXXVII.
parallelepipeds similarly described from them shall also be proportionals. And if the similar parallelepipeds fimi: larly described from four straight lines be proportionals, the straight lines shall be proportionals.
Let the four straight lines AB, CD, EF, GH be proportionals, viz. a6 AB to CD, so EF to GH ; and let the Gimilar parallelepipeds AK, CL, EM, GN be limilarly described from them. AK is to CL, as EM to GN.
Make AB, CD, O, P continual proportionals, as also EF, a 11. 6. GH, Q, R: Arid because as AB is to CD, so EF to GH; and
that CD is to O, as GH to Q, and Ở to P, as Q to R; there-0 11. 5.
Book XI. But let the folid AK be to the solid CL, as the solid EM to
the folid GN: The straight line AB is to CD, as EF to GH. € 27. II.
Take AB to CD, as EF to ST, and from ST describe a folid parallelepiped SV Gmilar and senilarly Gtuated to either of the solids EM, GN: And because AB is to CD, as EF to ST, and that from AB, CD the solid parallelepipeds AK, CL are fimilarly described ; and in like manner the folids EM, SV from the straight lines EF, 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 equalf to SV : But it is likewife fimilar and similarly situated to SV; therefore the planes which contain the solids GN, SV are fimilar and equal, and their homologous fides GH, ST equal to one another : And because 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 itraight lines, &c. Q. E. D.
a plane be perpendicular to another plane, and ut 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 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 66 AB Thail fall on AD.