DEFINITION. One plane is said to be perpendicular to another plane when every straight line drawn in one of the planes perpendicular to the common section of the two planes is perpendicular to the other plane, PROPOSITION XV. If a straight line be perpendicular to a plane, every plane which passes through it shall be perpendicular to that plane. Let AB be a straight line I to the plane HKLM, and CFG a plane passing through AB. In the common section FG take any point X, and through X in the plane CFG draw XY 1 to FG. PROPOSITION XVI. If two planes which cut one another be each of them perpendicular to a third plane, their common section shall be perpendicular to the same plane. Let the two planes ANB, CNF which cut one another be each i to the plane PQ. Then shall their common section SN be 1 to the plane PQ. For, if not, if possible, in plane ANB draw NX 1 to NB, and ... I to the plane PQ, and in the plane CNF draw NY Ito NF, and .. I to plane PQ, ... from the point N two straight lines have been drawn I to the plane PQ; which is impossible; (Prop. 11) .. SN is 1 to plane PQ. DEFINITION. A solid angle is that which is made by the meeting in one point of two or more plane angles, which are not in the same plane. PROPOSITION XVII. If a solid angle be contained by three plane angles any two of them are together greater than the third. N C Let the solid angle at A be contained by the three plane angles BAF, FAC, CAB. Then shall any two of them BAF, FAC be together greater than the third CAB. If either of the two angles BAF, FAC be = or > < CAB, the proposition is evidently true. But, if not, from 2 CAB cut off 2 BAN= 1 BAF; make AN= AF; through N draw BNC meeting AB, AC in B and C; join BF and CF. Then ::: BA, AF and 2 BAF are respectively = BA, AN and . BAN, . BF= BN. (1. 1) Again ::: BF, FC are together > BC and BF is = BN; 1. FC is > NC. is . FAC is > . NAC; (1. 19) .. Ls BAF, FAC are together > Ls BAN, NAC, i.e. >CAB. PROPOSITION XVIII. The plane angles which contain any solid angle are together less than four right angles. First let the solid angle at A be contained by three plane angles BAF, FAC, CAB. Then shall these angles be together less than four right 2 s. In AB, AC, AF take any points B, C, F, and join BC, CF, FB. Then since the solid < at B is contained by three plane angles; .(S ABC, ABF are together > < CBF. (Prop. 17) Similarly LS ACB, ACF are together >FCB, (Prop. 17) and ZS AFC, AFB > - BFC; (Prop. 17) i. LS ABC, ABF, ACB, ACF, AFC, AFB are together > the ls of a BCF, i.e. the es at the bases of the AS ABC, ACF, AFB are together > two right < S. (1. 24) Now all the zs of the As ABC, ACF, AFB are = six right 2 s; (1. 24) .. the es at the vertex A are < four right 2 s. Next, let the solid angle at A be contained by any number of plane angles. Then ls ACB, ACF are together > - BCF, (Prop. 17) SO LS AFC, AFG are together > LCFG: and so on. ... all the zs at the bases of the as ABC, ACF, &c. are together > the angles of the polygon BCF. Now all the zs of the as ABC, ACF, &c. are = twice as many right 2 s as there are As, (1. 24) and all the 2 s of the polygon BCF together with four right zs are equal to twice as many right 2 s as the figure has sides; (1. 30) ... all the zs of the as ABC, ACF, &c. are = the es of the polygon together with four right Zs; but the < s at the bases of the as ABC, ACF, &c. are > the zs of the polygon; .. the es at the vertex A are < four right 2 s. |