PROPOSITION IV. If a plane be perpendicular to the diameter of a sphere, at one of its extremities, it touches the sphere. Let the plane CD be perpendicular to AB, the diameter of the sphere ABG, at its extremity B; then CD touches the sphere in that point. For, let F be any other point in CD, and EF, FB, be drawn; the angle EBF is, by hypothesis, a right angle. Hence EF is greater than EB, and consequently F a point without the BE sphere. Thus the plane CD meets the sphere only in the point B, and therefore touches it. Cor. 1.-A sphere and a plane can touch one another only in one point. Cor. 2.-If a plane touch a sphere, the radius at the point of contact is perpendicular to it. Cor. 3.-If a plane touch a sphere, a perpendicular to it, at the point of contact, passes through the centre. Cor. 4.- If a plane touch a sphere, its line of common section, with the plane of any circle of the sphere passing through the point of contact, is a tangent to that circle. For this line of common section is in the plane of the circle, and it touches the circle. Cor. 5.-A tangent, to any circle of the sphere, is the common tangent of all the circles in whose plane it is. PROPOSITION V. In isosceles spherical triangles, the angles at the base are equal. Let ABC be a spherical triangle, having the side AB equal to the side AC; the spherical angles ABC and ACB are equal. Let D be the centre of the sphere ; join DB, DC, DA, and from A on the straight lines DB, DC, draw the perpendiculars AE, AF; and from the points É and F draw in the plane DBC the straight lines EG, FG, perpendicular to DB and DC, meeting one another in G; join AG. Because DE is at right angles to each of the straight lines AE, EG, it is at right angles to the plane AEG, which passes through AE, EG (So. Ge. I. 4); and, therefore, every plane that passes through DE is at right angles to the plane AEG (So. Ge. I. 17); wherefore, the plane DBC is at right angles to the plane AEG. For the same reason, the plane E B DBC is at right angles to the plane AFG; and therefore AG, the common section of the planes AFG, AEG, is at right angles (So. Ge. I. 18) to the plane DBC, and the angles AGE, AGF, are consequently right angles. But, since the arc AB is equal to the arc AC, the angle ADB is equal to the angle ABC. Therefore the triangles : ADE, ADF, have the angles EDA, FDA, equal, as also the angles AED, AFD, which are right angles ; and they have the side AD common; therefore the other sides are equal, viz. AE to AF (Pl. Ge. I. 26), and DE to DF. Again, because the angles AGE, AGF, are right angles, the squares on AG and GE are equal to the square of AE; and the squares of AG and GF to the square of AF. But the squares of AE and AF are equal; therefore the squares of AĞ and GE are equal to the squares of AG and GF; and taking away the common square of AG, the remaining squares of GE and GF are equal, and GE is therefore equal to GF. Wherefore, in the triangles AFG, AEG, the side GF is equal to the side GE, and AF has been proved to be equal to AE, and the base AG is common; therefore the angle AFG is equal to the angle AEG (Pl. Ge. I. 8). But the angle AFG is the angle which the plane ADC makes with the plane DBC (So. Ge. I. Def. 4), because FA and FG, which are drawn in these planes, are at right angles to DF, the common section of the planes. The angle AFG (Sp. Ge. Def. 7) is therefore equal to the spherical angle ACB; and, for the same reason, the angle AEG is equal to the spherical angle ABC. But the angles AFG, AEG, are equal. Therefore the spherical angles ACB, ABC, are also equal. PROPOSITION VI. If the angles at the base of a spherical triangle be equal, the triangle is isosceles. Let ABC be a spherical triangle having the angles ABC, ACB, equal to one another; the sides AC and AB are also equal. "Let D be the centre of the sphere; join DB, DC, DA, and from A on the straight lines DB, DC, draw the perpendiculars AE, AF; and from the points E and F, draw in the plane DBC the straight lines EG, FG, perpendicular to DB and DC, meeting one another in G; join AG. Then, it may be proved, as was done in the last proposition, that AG is at right angles to the plane BCD, and that therefore the angles AGF, AGE, are right angles, and also that the angles AFG, AEG, are equal to the angles which the planes DAC, DAB, make with the plane DBC. But because the spherical angles ACB, ABC, are equal, the angles which the planes DAC, DAB, make with the plane DBC, are equal (Sp. Ge. Def. 7), and therefore the angles AFG, AEG, are also equal. The triangles AGE, AĞF, have therefore two angles of the one equal to two angles of the other, and they have also the side AG common; wherefore they are equal, and the side AF is equal to the side AE. • Again, because the triangles ADF, ADE, are right angled at F and E, the squares of DF and FA are equal to the square of DA, that is, to the squares of A DE and EA; now, the square of AF is equal to the square of AE, therefore the square of DF is equal to the square of DE, and the side DF to the side DE. Therefore in the triangles DAF, DAE, L YG because DF is equal to DE, and DA D E B common, and also AF equal to AE, the angle ADF is equal to the angle ADE; therefore, also, the arcs AC and AB, which are the measures of the angles ADF and ADE, are equal to one another; and the triangle ABC is isosceles. в PROPOSITION VII. Any two sides of a spherical triangle are greater than the third. Let ABC be a spherical triangle, any two sides AB, BC, are greater than the third side AC. Let D be the centre of the sphere; join DA, DB, DC. The solid angle at D is contained by three plane angles ADB, ADC, BDC (So. Ge. II. 1); any two of which ADB, DK BDC, are greater than the third ADC; that is, any two sides AB, BC, of the spherical triangle ABC, are greater than the thir PROPOSITION VIII. The three sides of a spherical triangle are less than a circle. Let ABC be a spherical triangle as before, the three sides AB, BC, AC, are less than a circle. Let D be the centre of the sphere. The solid angle at D is contained by three plane angles BDA, BDC, ADC, which together are less than four right angles (So. Ge. II. 2); therefore the sides AB, BC, AC, together, will be less than four quadrants, that is, less than a circle. PROPOSITION IX. In a spherical triangle the greater angle is opposite to the greater side; and conversely. Let ABC be a spherical triangle, the greater angle A is opposed to the greater side BC. Let the angle BAD be made equal to the angle B, and then BD, DA, will be equal (Sp. Ge. 6), and therefore AD, DC, are equal to BC; but AD, DC, are greater than AC S (Sp. Ge. 7); therefore BC is greater than AC, that is, the greater angle A is opposite to the greater side BC. The converse is demonstrated as in Pl. Ge. I. 19. PROPOSITION X. In a spherical triangle, according as the sum of two of the sides is greater than a semicircle, equal to it, or less, the interior angle at the base is greater than the exterior and opposite angle at the base, equal to it, or less; and the sum of the two interior angles at the base greater than two right angles, equal to two right angles, or less than two right angles. Let ABC be a spherical triangle, of which the sides are AB and BC; produce the side AB and the base AC till they meet again in D; then, the arc ABD is a semicircle, and the spherical angles at A and D are equal, because each of them is the inclination of the circle ABD to the circle ACD. 1. If AB, BC, be equal to a semicircle, that is, to AD, BC will be equal to BD, and therefore (Sp. Ge. 5) the . angle D, or the angle A, will be equal to the angle BCD. 2. If AB, BC, together be / greater than a semicircle, that is, A greater than ABD, BC will be greater than BD; and therefore (Sp. Ge. 9) the angle D, that is, the angle A, is greater than the angle BCD. 3. In the same manner, it is shown if AB, BC, together be less than a semicircle, that the angle A is less than the angle BCD. And since the angles BCD, BCA, are equal to two right angles, if the angle A be greater than BCD, A and ACB together will be greater than two right angles. If A be equal to BCD, A and ACB together will be equal to two right angles; and if A be less that BCD, A and ACB will be less than two right angles. PROPOSITION XI. If the angular points of a spherical triangle be made the poles of three great circles, these three circles by their intersections will form a triangle which is said to be supplemental to the former; and the two triangles are such, that the sides of the one are the supplements of the arcs which measure. the angles of the other. |