alfo meet. Which is contrary to the Hypothefis. E G B PRO P. XVII. M F If two Right Lines ALB, Draw the Right Lines XVIII. If a Right Line F AB Angles to any Plane 2.6. Right Angles to the faid Plane CD. ત Draw fome Plane EF through AB, making the Section EG, with the Plane CD, from fome Point H whereof, draw HI in the 31. 1. Plane EF parall. to AB; then fhall HI be per- d 8. 11. pendicular to the Plane CD; and fo likewise fhall any other Perpendiculars. Therefore the Plane EF is perpendicular to the Plane CD, 4 def. 11. and by the fame Reason, any other Plane e drawn drawn thro' AB fhall be perpendicular to the faid Plane GH. are supposed at Because the Planes AB, CD, right Angles to the Plane GH, from Def. 4. 11. it is manifeft that from the Point F, a Perpend. may be drawn in each Plane AB, CD to the 13.11. Plane GH; and there can be but one only: therefore it fhall be their common Section. a be contain❜d un der three plané Angles BAD, DAC, BAC, any two of them, Chowsoever taken, are greater than the third: the Affertion is If the three Angles be equal, manifeft; if they be unequal, let BAC be the Note, That the first of the Letters expreffing a folid Angle, is at the Angular Point; but the laft Letter that denotes d Pyramid, is at the Vertex of the Pyramid. 1 great ь greatest: from which take a BAE=BAD; and c е fore the Ang. BAD CAD Q. E. D. ત b с 23. I. conft. 4. 1. 20. I. ax. 1. 25. I. BAC. 4 ax. I¿ h Sides of the fo lid Angle A a ny how cutting a Plane make the Triangular Figure BCD. and the Triangles ABC, ACD, ADB. Now let us call all the Angles of the Triangle BDC, X; and the Sum of the Angles at the Bafes of the Triangles Y. then X+ four right Angles YA. But becaufe 32. 1. (from the Angles at B,) the Ang. ABE&Schol. ABC is CBE; and fince the fame is true 32 1. of the Angles at C, D, it is manifeft that k YX. And confequently A¬ four right Angles. . E. D. PROP. XXII. k If there be three plane Angles A, B, HCI, any two of which together, are greater than the third, and the right Lines AD, AE, FB, &c. are equal; then it is poffible ro. make a Triangle of the Right Liner i I. 20. II. 5 ax.1. Lines DE, FG, HI, joining those equal Right Lines that form the Angles. с = A Triangle may be made of them, if any two taken together be greater than the third; but that the thing is fo, make the Ang. HCK = B, and CK CH, and draw HK, IK. then KH = FG. and because the Ang. KCI A; the Line KIDE. But KISHI+ HK (FG); therefore DE HI+FG. In like manner, we demonftrate that any two are greater than the third, and consequently a Triangle may be made of them. Q. Ê. D. a To make a folid Angle MHIK of three plane Angles A, B, C, whereof any two howsoever taken M * are greater than the third; but these three Angles* 21. 11. must be less than four right Angles. Make AD, AE, BE, BF, CF, CG equal to one another, and make the Triang. HKI of the three right Lines HI= DE, IK = EF, KH FG, about which defcribe a Circle LHKI. g f 22. I. h 5.4. 8. 1. 21. I. Now AD - HL, the Point L being the Centre; for if AD be, or HL, then will the Ang. AhorHLI. In like manner conft, & fhall B be, or HLK, and C, or KLI. Whence A+B+C are either equal k to four right Angles, or elfe exceed four right 13. 1. Angles, which is contrary to the Hypothefis ; therefore AD HL. 2 2 m 4 cor. Again, let AD2 be1 =HL+LM", and 1 Schol. let LM be perpend. to the Plane of the Circle 47. 1. HKI, and draw HM, KM, IM. Then be- 12. 11. caufe the Ang. HLM is " a right one; there- "3 def. 11. fore° MH2 HL2 + LM2 = PAD3. There- 47. 1. fore MH = AD. By the like Argument MK, & conft. MI, AD (that is, AE, EB, &c.) are equal; whence fince HM = AD, and MI AE, and 9 8. 1. DEP HI; = HI; therefore is the Ang. A HMI; and in like manner the Ang. IMK B, and the Ang. HMK = C C. Whence the folid Angle at M is made of three given Plane Angles. Q. E. F. |