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 1 to the plane HKLM, Then shall CFG be 1 to the plane HKL. In the common section FG take any point X, and through X in the plane CFG draw XY1 to FG. Then AB is 1 to plane HKL,.. AB is 1 to FG; (Prop. 7) (Def.) .. YX is to plane HKL; .. the plane CFG is 1 to the plane HKL. 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 1 to the plane PQ. Then shall their common section SN be to the plane PQ. For, if not, if possible, in plane ANB draw NX1 to NB, and.. to the plane PQ, and in the plane CNF draw NY 1 to NF, and.. 1 to plane PQ, ... from the point N two straight lines have been drawn 1 to the plane PQ; which is impossible; (Prop. 11) 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. 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 4 CAB cut off ▲ BAN=L BAF; make AN = AF; through N draw BNC meeting AB, AC in B and C; join BF and CF. Then BA, AF and BAF are respectively = BA, AN and BAN, .. BF = BN. (1. 1) Again . BF, FC are together > BC and BF is = BN; .. FC is > NC. Also AF, AC are respectively = AN, AC; ..L FAC is > L NAC; (1. 19) .. LS BAF, FAC are together > < s BAN, NAC, i. e. > L 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 rights. In AB, AC, AF take any points B, C, F, and join BC, CF, FB. angles; .. LS ABC, ABF are together > CBF. (Prop. 17) Similarlys ACB, ACF are together > FCB, (Prop. 17) and S AFC, AFB > L BFC; (Prop. 17) .. LS ABC, ABF, ACB, ACF, AFC, AFB are to gether the 4s of a BCF, i. e. the 4s at the bases of the AS ABC, ACF, AFB are together two rights. (I. 24) six Now all the 4s of the as ABC, ACF, AFB are = rights; .. thes at the vertex A are < four rights. (1. 24) Next, let the solid angle at A be contained by any number of plane angles. B H G F Let AB, AC, &c. be cut by a plane in the points B, C, F, &c. and join BC, CF, FG, &c. Then s ACB, ACF are together > BCF, SOLS AFC, AFG are together > (Prop. 17) CFG: and so on. ... all the 4s at the bases of the as ABC, ACF, &c. are together the angles of the polygon BCF Now all the 4s of the as ABC, ACF, &c. are as many rights as there are as, = twice (1. 24) and all the 4s of the polygon BCF together with four rights are equal to twice as many rights as the figure has sides; ... all the 4s of the as ABC, ACF, &c. are = of the polygon together with four right s; (I. 30) the s but the 4s at the bases of the as ABC, ACF, &c. are >thes of the polygon; .. thes at the vertex A are < four rights. |