but BAC is not less than either of the angles DAB and DAC; therefore 5. BAC with either of them is greater than the third. Wherefore, if a solid angle, &c. Q.E.D. PROPOSITION 21.-Theorem. The plane angles by which every solid angle is contained, are together less than four right angles. First, let the solid angle at A be contained by three plane angles BAC, CAD, and DAB. Then these three together are less than four right angles. D B Construction. Take in each of the straight lines AB, AC, and AD, any points B, C, and D; and join BC, CD, and DB. Demonstration. Because the solid angle at B is contained by the three plane angles CBA, ABD, and DBC; any two of them are greater than the third (XI. 20); therefore 1. The angles CBA and ABD are greater than the angle DBC. For the same reason, the angles BCA and ACD are greater than the angle DCB; and the angles CDA and ADB greater than the angle BPC; therefore 2. The six angles CBA, ABD, BCA, ACD, CDA, and ADB are greater than the three angles DBC, BCD, and CDB; but the angles DBC, BCD, and CDB are equal to two right angles (I. 32); therefore 3. The six angles CBA, ABD, BCA, ACD, CDA, and ADB are greater than two right angles. Because the three angles of each of the triangles ABC, ACD, and ADB are equal to two right angles ; therefore 4. The nine angles of these three triangles-viz., the angles CBA, BAC, ACB, ACD, CDA, DÅC, ADB, DRA, and BAD are equal to six right angles ; but of these, the six angles CBA, ACB, ACD, CDA, ADB, and DBA are greater than two right angles ; therefore 5. The remaining three angles BAC, CAD, and DAB, which contain the solid angle at A, are less than four right angles. Next, let the solid angle at A be contained by any number of plane angles BAC, CAD, DAE, EAF, and FAB. Then these angles are together less than four right angles, B E Construction. Let the plane angles which form the solid angle be cut by a plane at any distance from its vertex, and let the common sections of it with those plane angles be BC, CD, DE, EF, and FB. Demonstration. Because the solid angle at B is contained by three plane angles CBA, ABF, and FBC, of which any two are greater than the third (XI. 20); therefore 1. The angles CBA and ABF are greater than the angle FBC. For the same reason, the two plane angles at each of the points C, D, E, and F-viz., those angles which are at the bases of the triangles having the common vertex A, are greater than the third angle at the same point, which is one of the angles of the polygon BCDEF; therefore 2. All the angles at the bases of the triangles are together greater than all the angles of the polygon. Because all the angles of the triangles are together equal to twice as many right angles as there are triangles (I. 32);-that is, as there are sides in the polygon BCDEF; and all the angles of the polygon, together with four right angles, are likewise equal to twice as many right angles as there are sides in the polygon (I. 32, Cor. 1); therefore 3. All the angles of the triangles are equal to all the angles of the polygon together with four right angles (I. Ax. 1); but all the angles at the bases of the triangles are greater than all the angles of the polygon, as has been proved ; therefore 4. The remaining angles of the triangles-viz., those at the vertex, which contain the solid angle at A, are less than four right angles. Wherefore, the plane angles, &c. Q.E.D. BOOK XII. (PROPOSITIONS 1 AND 2.) LEMMA.* t [This lemma, which is usually inserted here, is the first proposition of Euclid, Book X. It is necessary to the understanding of some of the propositions of Book XII.] If from the greater of two unequal magnitudes of the same kind, there be taken more than its half, and from the remainder more than its half; and so on: there will at length remain a magnitude less than the least of the proposed magnitudes. Let AB and C be two unequal magnitudes, of the same kind, 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 ; Then there will at length remain a magnitude less than C. Construction. 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; also let DE be divided into parts DF, FG, and GE, each equal to C. From AB take BH greater than its half, and * A Lemma is a proposition of no importance in itself, merely introduced for the purpose of demonstrating some other proposition. from the remainder AH take HK greater than its half, and so on, until there be as many divisions in Å B, as there are in DE; and let the divisions in AB be AK, KH, and HB; and the divisions in DE be DF, FG, and GE. Demonstration. Because DE is greater than AB, and EG taken from DE is not greater than its half, but BH taken from AB is greater than its half; therefore 1. The remainder GD is greater than the remainder HA. Again, because GD is greater than HA, and that GF is not greater than the half of GD, but HK is greater than the half of HA; therefore 2. The remainder FD is greater than the remainder AK; but FD is equal to C; therefore 3. C is greater than AK; that is, AK is less than C. Wherefore, if from the greater, &c. Q.E.D. Corollary. If only the halves be taken away, the same thing may in the same way be demonstrated. PROPOSITION 1.-Theorem. Similar polygons inscribed in circles, are to one another as the squares on their diameters. Let ABCDE and FGHKL be two circles having the similar polygons ABCDE and FGHKL inscribed in them; and let BM and GN be the diameters of the circles. Then the polygon ABCDE is to the polygon FGHKL as the square on BM is to the square on GN. Construction. Join BE, AM, GL, and FN. |