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PROP. XXXV. THEOR.
IF, from the vertices of two equal plane angles, there be drawn two straight lines elevated above the planes in which the angles are, and containing equal angles with the sides of those angles, each to each; and if in the lines above the planes there be taken any points, and from them perpendiculars be drawn to the planes in which the first named angles are: And from the points in which they meet the planes, straight lines be drawn to the vertices of the angles first named; these straight lines shall contain equal angles with the straight lines which are above the planes of the angles. *
Let BAC, EDF be two equal plane angles ; and from the points A, D let the straight lines AG, DM be elevated above the planes of the angles, making equal angles with their sides, each to each, viz. the angle GAB equal to the angle MDE, and GAC to MDF: and in AG, DM let any points G, M be taken, and from them let perpendiculars GL, MN be drawn to the planes
BAC, EDF, meeting these planes in the points L, N, and join LA, ND: the angle GAL is equal to the angle MDN.
Make AH equal to DM, and through H draw HK parallel to GL. But GL is perpendicular to the plane BAC; wherefore HK is perpendicular (8. 11.) to the same plane : from the points K, N to the straight lines AB, AC, DE, DF, draw perpendiculars KB, KC, NË, NF; and join HB, BC, ME, EF :
* See Note.
Because HK is perpendicular to the plane BAC, the plane HBK which passes through HK is at right angles (18. 11.) to the plane BAC: and AB is drawn in the plane BAC at right angles to the common section BK of the two planes ; therefore AB is perpendicular (4. def. 11.) to the plane HBK, and makes right angles (3. def. 11.) 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 side, opposite to one of the equal angles in each, viz. HA equal to DM; therefore the remaining sides are equal (26. 1.) each to each : wherefore AB is equal to DE. In the same manner, if HC and MF be joined, it may be demonstrated that AC is equal to DF; therefore, since 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 equal (4. 1.) 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 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 square of DM: but the squares of AK, KH are equal to the square (47. 1.) of AH, because AKH is a right angle : and the squares of DN, NM are equal 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 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, and the base HK to the base MN, as has been proved ; therefore the angle HAK is equal (8. 1.) to the angle MDN. Q. E. D.
Cor. From this it is manifest, that if, from the vertices of two equal plane angles, there be elevated two equal straigh 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 solid 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 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 shown 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 MN (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.
If three straight lines be proportionals, the solid parallelopiped described from all three as its sides, is equal to the equilateral parallelopiped 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 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 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 parallelopiped DH. Make LK equal to A, and at the point K in the straight line LK make (26. 11.) a solid angle contained by the three plane angles LKM, MKN, NKL, equal to the angles EDF, FDG, GĎE,
с each to each; and make KN equal to B, and KM equal to C; and complete the_solid parallelopiped 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 DF to KM ; that is, the sides about the equal angles are reciprocally proportional ; therefore the parallelogram LM is equal (14. 6.) 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 sides; therefore the perpendiculars from the points G, N, to the planes EDF, LKM are equal
. See Note.
(CoR. 35. 11.) to one another: therefore the solids KO, DH of the same altitude; and they are upon equal bases LM, EF, and therefore they are equal (31. 11.) to one another : but the solid KO is described from the three straight lines A, B, C, and the solid DH from the straight line B. If therefore three straight lines, &c. Q. E. D.
PROP. XXXVII. THEOR.
If four straight lines be proportionals, the similar solid parallelopipeds similarly described from them shall also be proportionals. And if the similar parallelopipeds similarİy 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. as AB to CD, so EF to GH; and let the similar parallelopipeds AK, CL, EM, GN be similarly described from them. AK is to CL, as EM to GN.
Make(11.6.) AB,CD, O, P continual proportionals, as also EF, GH, Q, R; and because as AB is to CD, so EF to GH; and
Ꭱ ; that CD is (11.5.) to 0, as GH to Q, and 0 to P, as Q to R; therefore, ex æquali (22. 5.), AB is to P, as EF to R: but as AB to P, so (Cor. 33. 11.) is the solid AK to the solid Chand as EF to R, so (Cor. 33. 11.) is the solid EM to the solid GN; therefore (11. 5.) as the solid AK to the solid CL, so is the solid EM to the solid GN.
• See Note.