Let the two planes BA, BC be each of them perpendicular to a third plane, and let BD be the common section of the planes BA, BC: BD shall be perpendicular to the third plane. For, if not, from the point D, draw in the plane BA, the straight line DE at right angles to AD, the common section of the plane BA with the third plane; [I. 11. and from the point D, draw in the plane BC, the straight line DF at right angles to CD, the common section of the plane BC with the third plane. [I. 11. Then, because the plane BA is perpendicular to the third plane, [Hypothesis. and DE is drawn in the plane BA at right angles to AD their common section; [Construction. therefore DE is perpendicular to the third plane. (XI. Def. 4. In the same manner it may be shewn that DF is perpendicular to the third plane. Therefore from the point D two straight lines are at right angles to the third plane, on the same side of it; which is impossible. [XI. 13. Therefore from the point D, there cannot be any straight line at right angles to the third plane, except BD the common section of the planes BA, BC; therefore BD is perpendicular to the third plane. Wherefore, if two planes &c. Q.E.D. PROPOSITION 20. THEOREM. 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 BAC, CAD, DAB: any two of them shall be together greater than the third. If the angles BAC, CAD, DAB be all equal, it is evident that any two of them are greater than the third. If they are not all equal, let BAC be that angle which is not less than either of the other two, and is greater than one of them, BAD. At the point A, in the straight line BA, make, in the plane which passes through BA, AC, the angle BAE equal to the angle BAD; [I. 23. make AE equal to AD; [I. 3. through E draw BEC cutting AB, AC at the points B, C; and join DB, DC. Then, because AD is equal to AE, [Construction and AB is common to the two triangles BAD, BAE, the two sides BA, AD are equal to the two sides BA, AE, each to each; and the angle BAD is equal to the angle BAE; [Constr. therefore the base BD is equal to the base BE. [I. 4. And because BD, DC are together greater than [I. 20. and one of them BD has been shewn equal to BE a part of BC, therefore the other DC is greater than the remaining part EC. And because AD is equal to AE, [Construction. and AC is common to the two triangles DAC, EAC, but the base DC is greater than the base EC; therefore the angle DAC is greater than the angle EAC. [I. 25. And, by construction, the angle BAD is equal to the angle BAE; therefore the angles BAD, DAC are together greater than the angles BAE, EAC, that is, than the angle BAC. But the angle BAC is not less than either of the angles BAD, DAC; therefore the angle BAC together with either of the other angles is greater than the third. Wherefore, if a solid angle &c. Q.E.D. PROPOSITION 21. THEOREM, Every solid angle is contained by plane angles, which are together less than four right angles. First let the solid angle at A be contained by three plane angles BAC, CAD, DAB: these three shall be together less than four right angles. In the straight lines AB, AC, AD take any points B, C, D, and join BC, CD, DB. Then, because the solid angle at B is contained by the three plane angles CBA, ABD, DBC, any two of them are together greater than the third, (XI. 20. therefore the angles CBA, ABD are together greater than the angle DBC. For the same reason, the angles BCA, ACD are together greater than the angle DCB, and the angles CDA, ADB are together greater than the angle BDČ. Therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are together greater than the three angles DBC, DCB, BDC; but the three angles DBC, DCB, BDC are together equal to two right angles. [I. 32 Therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are together greater than two right angles. And, because the three angles of each of the triangles ABC, ACD, ADB are together equal to two right angles, [I. 32. therefore the nine angles of these triangles, namely, the angles CBA, BAC, ACB, ACD, CDĂ, CAD, ADB, DBA, DAB are equal to six right angles ; and of these, the six angles CBA, ACB, ACD, CDA, ADB, DBA are greater than two right angles, therefore the remaining three angles BAC, CAD, DAB, which contain the solid angle at X, are together less than four right angles. Next, let the solid angle at A be contained by any number of plane angles BAC, CAD, DAE, EAF, FAB: these shall be together less than four right angles. Let the planes in which the angles are, be cut by a plane, and let the common sections of it with those planes be BC, CD, DE, EF, FB. Then, because the solid angle at angles CBA, ABF, FBC, any two of them are together greater than the third, [XI. 20. therefore the angles CBA, ABF are together greater than the angle FBC. For the same reason, at each of the points C, D, E, F, the two plane angles which are at the bases of the triangles having the common vertex A, are together greater than the third angle at the same point, which is one of the angles of the polygon BCDEF. Therefore all the angles at the bases of the triangles are together greater than all the angles of the polygon. Now all the angles of the triangles are together equal to twice as many right angles as there are triangles, that is, as there are sides in the polygon BCDEF; [I. 32. angles, are also equal to twice as many right angles as there are sides in the polygon; [I. 32, Corollary 1. therefore all the angles of the triangles are equal to all the angles of the polygon, together with four right angles. [A.. 1. But it has been shewn that all the angles at the bases of the triangles are together greater than all the angles of the polygon ; therefore the remaining angles of the triangles, namely, together less than four right angles. Wherefore, every solid angle &c. Q.E.D. BOOK XII. LEMMA. If from the greater of two unequal magnitudes there be taken more than its half, and from the remainder more than its half, and so on, there shall at length remain a magnitude less than the smaller of the proposed magnitudes. Let AB and C be two unequal magnitudes, of which AB is the greater: if from AB there be taken more than its half, and from the remainder more than its half, and so on, there shall at length remain a magnitude less than C. For C may be multiplied so as at length to become greater than AB. Let it be so multiplied, and let DE its multiple be greater than AB, and let DE K be divided into DF, FG, GE, each equal 7 to C. From AB take BH, greater than its half, and from the remainder AH take HK greater than its half, and so on, until there be as many divisions in AB as in DE; and let the divisions in AB be AK, KH, HB, and the divisions in DE be DF, FG, GE. Then, because DE is greater than AB: . and that EG taken from DE is not greater than its half; but BH taken from AB is greater than its half ; therefore the remainder DG is greater than the remainder AH. |