For, conceive a small circle to be drawn through A; B, D and C, and let P be its pole; draw arcs of great circles from P to A, B, and C: these arcs will be equal (D. 7). Draw the arc of a great circle FQ, making the angle DFQ equal to ACP, and lay off off on it, FQ equal to CP; draw arcs of great circles QD and QE Q P B equal to AC, by hypothesis; the side FQ equal to PC by construction, and the angle DFQ equal to ACP, by construction : hence (P. VIII.), the side DQ is equal to AP, the angle FDQ to PAC, and the angle FQD to APC. Now, because the triangles QFD and PAC are isosceles and equal in all their parts, they may be placed so as to coincide throughout, the base FD falling on AC, DQ on CP, and FQ on AP: hence, they are equal in area. If we take from the angle DFE the angle DFQ, and from the angle ACB the angle ACP, the remaining angles QFE and PCB, will be equal. In the triangles FQE and PCB, we have the side QF equal to PC, by construction, the side FE equal to BC, by hypothesis, and the angle QFE equal to PCB, from what has just been shown: hence, the triangles are equal in all their parts, and being isosceles, they may be placed so as to coincide throughout, the side QE falling on PC, and the side QF on PB; these triangles are, therefore, equal in area. In the triangles QDE and PAB, we have the sides QD, QE, PA, and PB, all equal, and the angle DQE equal to APB, because they are the sums of equal angles : hence, the triangles are equal in all their parts, and because they are isosceles, they may be so placed as to coincide throughout, the side QD falling on PB, and the side QE on PA; these triangles are, therefore, equal in area. Hence, the sum of the triangles QFD and QFE, is equal to the sum of the triangles PAC and PBC. If from the former sum we take away the triangle QDE, there will remain the triangle DFE; and if from the latter sum take away the triangle PAB, there will remain the triangle ABC: hence, the triangles ABC and DEF are equal in area; which was to be proved. Scholium. If the point P falls within the triangle ABC, the point will fall within the triangle DEF. In this case, the triangle DEF is equal to the sum of the triangles QFD, QFE, and QDE, and the triangle ABC is equal to the sum of the equal triangles PAC, PBC, and PAB; the proposition, therefore, still holds good. PROPOSITION XVII. THEOREM. If the circumferences of two great circles intersect intersect on the surface of a hemisphere, the sum of the opposite triangles thus formed, is equal to a lune whose angle is equal to that formed by the circles. Let the circumferences AOB, COD, intersect on the surface of a hemisphere: then will the sum of the opposite triangles AOC, BOD, be equal to the lune whose angle is BOD. For, produce the arcs OB, OD, on the other hemisphere, till they meet at N. Now, since AOB and OBN A are semi-circumferences, if we take away the common part OB, we shall have BN equal to 40. For a like rea son, we have DN equal to CO, and BD equal to AC: hence, the two triangles AOC, BDN, Schoum. It is evident that the two spherical pyramids, which have the triangles AOC, BOD, for for bases, are together equal to the spherical wedge whose angle is BOD. PROPOSITION XVIII. THEOREM. The area of a spherical triangle is equal to its spherical excess multiplied by a tri-rectangular triangle. Let ABC be a spherical triangle: then will its surface be equal to (A+B+C - 2) × T For, produce its sides till they meet the great circle DEFG, drawn at pleasure, without the triangle. By the last theorem, the two triangles ADE, AGH, are together equal to the lune whose angle is A; but the area of this lune is equal to 24 × T (P. XV., C. 2): hence, the sum of the triangles ADE and AGH, is equal to 24 × T. In like manner, it may be shown that the sum of the triangles BFG and BID, is equal to 2B × T, and that the sum of the triangles CIH and CFE, is equal to 2C × T. But the sum of these six triangles exceeds the hemis phere, or four times T, by twice the triangle ABC. We shall therefore have, 2 × area ABC = 2A × T + 2B × T + 2C × T − 4T; or, by reducing and factoring, area ABC = (A + B + C − 2) × T ; which was to be proced. - Scholium 1. The same relation which exists between the spherical triangle ABC, and the tri-rectangular triangle, exists also between the spherical pyramid which has ABC for its base, and the tri-rectangular pyramid. The triedral angle of the pyramid is to the triedral angle of the trirectangular pyramid, as the triangle ABC to the tri-rectangular triangle. From these relations, the following consequences are deduced: 1o. Triangular spherical pyramids are to each other as their bases; and since a polygonal pyramid may always be divided into triangular pyramids, it follows that any two spherical pyramids are to each other as their bases. 2o. Polyedral angles at the centre of the same, or of equal spheres, are to each other as the spherical polygons. intercepted by their faces. Scholium 2. A triedral angle whose faces are perpendicular to each other, is called a right triedral angle; and if the vertex be at the centre of a sphere, its faces will intercept a tri-rectangular triangle. The right triedral angle is taken as the unit of polyedral angles, and the tri-rectangular spherical triangle is taken as its measure. If the vertex of a polyedral angle be taken as the centre of a sphere, the portion of the surface intercepted by its faces will be the measure of the polyedral angle, a tri-rectangular triangle of he same sphere, being the unit. The area of a spherical polygon is equal to its spherical excess multiplied by the tri-rectangular triangle. Let ABCDE be a spherical polygon, the sum of whose angles is S, and the number of whose sides is n: then will its area be equal to the sum of the areas of all the triangles, or the area of the polygon, is equal to the sum of all the angles of the triangles, or the sum of the angles of the polygon diminished by 2(n-2) into the tri-rectangular triangle; or, area ABCDE = [S2(n-2)] × T; whence, by reduction, |