PROPOSITION I. THEOREM.

Every section of a sphere, made by 1 plane, is a circle

Let ABD be a section, made by a plane, in a sphere whose center is C. From the point C draw CE perpendicu- A lar to the plane ABD; and draw lines CA, CB, CD, &c., to different points of the curve ABD which bounds the section.

The oblique lines CA, CB, CD are equal, because they are radii of the

sphere; therefore they are equally distant from the perpen. dicular CE (Prop. V., Cor., B. VII.). Hence all the lines EA, EB, ED are equal; and, consequently, the section ABD is a circle, of which E is the center. Therefore, every section, &c.

Cor. 1. If the section passes through the center of the sphere, its radius will be the radius of the sphere; hence all great circles of a sphere are equal to each other.

Cor. 2. Two great circles always bisect each other; for, since they have the same center, their common section is a diameter of both, and therefore bisects both.

Cor. 3. Every great circle divides the sphere and its surface into two equal parts. For if the two hemispheres are separated and applied to each other, base to base, with their convexities turned the same way, the two surfaces must coincide; otherwise there would be points in these surfaces unequally distant from the center.

Cor. 4. The center of a small circle, and that of the sphere, are in a straight line perpendicular to the plane of the small circle.

Cor. 5. The circle which is furthest from the center is the least; for the greater the distance CE, the less is the chord AB, which is the diameter of the small circle ABD.

Cor. 6. An arc of a great circle may be made to pass through any two points on the surface of a sphere; for the two given points, together with the center of the sphere. make three points which are necessary to determine the position of a plane. If, however, the two given points were situated at the extremities of a diameter, these two points and the center would then be in one straight line, and any num ber of great circles might be made to pass through them.

PROPOSITION II. THEOREM.

Any two sides of a spherical triangle are greater than the third.

Let ABC be a spherical triangle; any two sides as, AB, BC, are together greater A than the third side AC.

B

Let D be the center of the sphere; and join AD, BD, CD. Conceive the planes ADB, BDC, CDA to be drawn, forming at solid angle at D. The angles ADB, BDC, CDA will be measured by AB, BC, CA, the sides of the spherical triangle. But when a solid angle is formed by three plane angles, the sum of any two of them is greater than the third (Prop. XVII., B. VII.); hence any two of the arcs AB, BC, CA must b greater than the third. Therefore, any two sides, &c.

D

PROPOSITION III. THEOREM.

The shortest path from one point to another on the surface of a sphere, is the arc of a great circle joining the two given points.

Let A and B be any two points on the surface of a sphere, and let ADB be the arc of a great circle which joins them; then will the line ADB be the shortest path from A to B on the surface of the sphere.

For, if possible, let the shortest path from A to B pass through C, a point situated out of the arc of a great circle ADB. Draw AC, CB, arcs of great circles, and take BD equal to BC.

B

D

By the preceding theorem, the arc ADB is less than AC+ CB. Subtracting the equal arcs BD and BC, there will remain AD less than AC. Now the shortest path from B to C, whether it be an arc of a great circle, or some other line, is equal to the shortest path from B to D; for, by revolving BC around B, the point C may be made to coincide with D, and thus the shortest path from B to C must coincide with the shortest path from B to D. But the shortest path from A to B was supposed to pass through C; hence the shortest path from A to C, can not be greater than the shortest path from A to D.

Now the arc AD has been proved to be less than AC; and therefore if AC be revolved about A until the point C falls on the arc ADB, the point C will fall between D and B. Hence the shortest path from C to A must be greater thar the shortest path from D to A; but it has just been proved not to be greater, which is absurd. Consequently, no point of the shortest path from A to B, can be out of the arc of a great circle ADB. Therefore, the shortest path, &c.

PROPOSITION IV. THEOREM.

The sum of the sides of a spherical polygon, is less than the circumference of a great circle.

Let ABCD be any spherical polygon; then will the sum of the sides AB, BC, CD, DA be less than the circumference of a great circle.

Let E be the center of the sphere, and join AE, BE, CE, DE. The solid angle at E is contained by the plane angles AEB, BEC, CED, DEA, which together are less than four right angles (Prop. XVIII., B. VII.). Hence the sides AB, BC, CD, DA, which are the measures of these angles, are

A

E

D

C

B

together less than four quadrants described with the radius AE; that is, than the circumference of a great circle. Therefore, the sum of the sides, &c.

PROPOSITION V. THEOREM.

The extremities of a diameter of a sphere, are the poles of all circles perpendicular to that diameter.

Let AB be a diameter perpendicular to CDE, a great circle of a sphere, and also to the small circle FGH; then will A and B, the extremities of the diameter, be the poles of both these circles.

For, because AB is perpendicular to the plane CDE, it is perpendicular to every straight line CI, DI, EI, &c., drawn through its foot in the plane ;

C

hence all the arcs AC, AD. AE, &c., are quarters of the cu

cumference. So, also, the arcs BC, BD, BE, &c., are quarters of the circumference; hence the points A and B are each equally distant from all the points of the circumference CDE; they are, therefore, the poles of that circumference (Def. 5).

Secondly. Because the radius AI is perpendicular to the plane of the circle FGH, it passes through K, the center of that circle (Prop. I., Cor. 4). Hence, if we draw the oblique lines AF, AG, AH, these lines will be equally distant from the perpendicular AK, and will be equal to each other (Prop. V., B. VII.). But since the chords AF, AG, AH are equal, the arcs are equal; hence the point A is a pole of the small circle FGH; and in the same manner it may be proved that B is the other pole.

Cor. 1. The arc of a great circle AD, drawn from the pole to the circumference of another great circle CDE, is a quadrant; and this quadrant is perpendicular to the arc CD. For, because AI is perpendicular to the plane CDI, every plane ADB which passes through the line AI is perpendicular to the plane CDI (Prop. VI., B. VII.); therefore the an. gle contained by these planes, or the angle ADC (Def. 6), is a right angle.

Cor. 2. If it is required to find the pole of the arc CD, draw the indefinite arc DA perpendicular to CD, and take DA equal to a quadrant; the point A will be one of the poles of the arc CD. Or, at each of the extremities C and D, draw the arcs CA and DA perpendicular to CD; the point of inter section of these arcs will be the pole required.

Cor. 3. Conversely, if the distance of the point A from each of the points C and D is equal to a quadrant, the point A will be the pole of the arc CD); and the angles ACD, ADC will be right angles.

For, let I be the center of the sphere, and draw the radii AI, CI, DI. Because the angles AIC, AID are right angles, the line AI is perpendicular to the two lines CI, DI; it is, therefore, perpendicular to their plane (Prop. IV., B. VII.). Hence the point A is the pole of the arc CD (Prop. V.); and therefore the angles ACD, ADC are right angles (Cor. 1).

Scholium. Circles may be drawn upon the surface of a sphere, with the same ease as upon a plane surface. Thus, by revolving the arc AF around the point A, the point F will describe the small circle FGH; and if we revolve the quadrant AC around the point A, the extremity C will describe the great circle CDE.

If it is required to produce the arc CD, or if it is required to draw an arc of a great circle through the two points C and D, then from the points C and D as centers, with a radius

equal to a quadrant, describe two arcs intersecting each other in A. The point A will be the pole of the arc CD; and, therefore, if, from A as a center, with a radius equal to a quadrant, we describe a circle CDE, it will be a great circle passing through C and D.

If it is required to let fall a perpendicular from any point G upon the arc CD; produce CD to L, making DL equal to a quadrant; then from the pole L, with the radius DL, describe the arc GD; it will be perpendicular to CD.

PROPOSITION VI. THEOREM.

A plane, perpendicular to a diameter at its extremity, touches the sphere.

Let ADB be a plane perpendicular A to the diameter DC at its extremity; then the plane ADB touches the sphere.

Let E be any point in the plane ADB, and join DE, CE. Because CD is perpendicular to the plane ADB, it is perpendicular to the line AB (Def.

B

1, B. VII.); hence the angle CDE is a right angle, and the line CE is greater than CD. Consequently, the point E lies without the sphere. Hence the plane ADB has only the point D in common with the sphere; it therefore touches the sphere (Def. 4). Therefore, a plane, &c.

Cor. In the same manner, it may be proved that two spheres touch each other, when the distance between their centers is equal to the sum or difference of their radii; in which case, the centers and the point of contact lie in one straight line.

PROPOSITION VII. THEOREM.

The angle formed by two arcs of great circles, is equal to the angle formed by the tangents of those arcs at the point of their intersection; and is measured by the arc of a great circle described from its vertex as a pole, and included between its sides.

Let BAD be an angle formed by two arcs of great circles; then will it be equal to the angle EAF formed by the tan