gles or the polygon, together with four right angles, are equa to twice as many right angles as the figure has sides (Prop. XXVIII., B. I.); hence all the angles of the triangles are equal to all the angles of the polygon, together with four right angles. But it has been proved that the angles at the cases of the triangles, are greater than the angles of the polygon. Hence the remaining angles of the triangles, viz., those which contain the solid angle at A, are less than four right angles. Therefore, the plane angles, &c.

Scholium. This demonstration supposes that the solid angle is convex; that is, that the plane of neither of the faces, if produced, would cut the solid angle. If it were otherwise, the sum of the plane angles would no longer be limited, and might be of any magnitude.

PROPOSITION XIX. THEOREM.

If two solid angles are contained by three plane angles which are equal, each to each, the planes of the equal angles will ba •qually inclined to each other.

to the inclination of the planes abc, abd.

D b

a

In the line AC, the common section of the planes ABC, ACD, take any point C; and through C let a plane BCE pass perpendicular to AB, and another plane CDE perpendicular to AD. Also, take ac equal to AC; and through c let a plane bce pass perpendicular to ab, and another plar.e cde perpendicular to ad.

Now, since the line AB is perpendicular to the plane BCE, it is perpendicular to every straight line which it meets in that plane; hence ABC and ABE are right angles. For the same reason abc and abe are right angles. Now, in the tri angles ABC, abc, the angle BAC is, by hypothesis, equal to bac, and the angles ABC, abc are right angles; therefore the angles ACB, acb are equal. But the side AC was made equal to the side ac; hence the two triangles are equal (Prop. VU., B. I); that is. the side AB is equal to ab, and BC

to be. In the same manner, it may be proved that AD is equal to ad, and CD to cd:

We can now prove that the quadrilateral ABED is equal to the quadrilateral abed. For, let the angle BAD be placed upon the equal angle bad, then the point B will fall upon the point b, and the point D upon the point d; because AB is equal to ab, and AD to ad. At the same time. BE, which is perpendicular to AB, will fall upon be, which is perpendicu lar to ab; and for a similar reason DE will fall upon de Hence the point E will fall upon e, and we shall have BE equal to be, and DE equal to de.

Now, since the plane BCE is perpendicular to the line AB, it is perpendicular to the plane ABD which passes through AB (Prop. VI.). For the same reason CDE is perpendicular to the same plane; hence CE, their common section, is perpendicular to the plane ABD (Prop. VIII.). In the same manner, it may be proved that ce is perpendicular to the plane abd. Now, in the triangles BCE, bce, the angles BEC, bec are right angles, the hypothenuse BC is equal to the hypothenuse bc, and the side BE is equal to be; hence the two triangles are equal, and the angle CBE is equal to the angle cbe. But the angle CBE is the inclination of the planes ABC, ABD (Def. 4); and the angle cbe is the inclination of the planes abc, abd; hence these planes are qually inclined to each other.

We must, however, observe that the angle CBE is not, properly speaking, the inclination of the planes ABC, ABD, except when the perpendicular CE falls upon the same side of AB as AD does. If it fall upon the other side of AB, then the angle between the two planes will be obtuse, and this angle, together with the angle B of the triangle CBE, will make two right angles. But in this case, the angle between the two planes abc, abd will also be obtuse, and this angle, together with the angle b of the triangle cbe, will also make two right angles. And, since the angle B is always equal to the angle b, the inclination of the two planes ABC, ABD will always be equal to that of the planes abc, abd. Therefore, if wo solid angles, &c.

Scholium. If two solid angles are contained by three plane angles which are equal, each to each, and similarly situated, the angles will be equal, and will coincide when applied the one to the other. For we have proved that the quadrilateral ABED will coincide with its equal abed Now, because the triangle BCE is equal to the triangle bce, the line CE, which is perpendicular to the plane ABED, is equal to the line ce, which is perpendicular to the plane abed. And since only one perpendicular can be drawn to a plane

gles do not admit of superposition, unless the three equal plane angles are similarly situated in both cases. For if the perpendiculars CE, ce lay on opposite sides of the planes ABED abed, the two solid angles could not be made to coincide Nevertheless, the Proposition will always hold true, that the planes containing the equal angles are equally inclined to each other.

BOOK VIII.

POLYEDRONS.

Definitions.

་

. A polyedron is a solid included by any numbe of planes which are called its faces. If the solia have only four faces which is the least number possible, it is called a tetraedron, if six faces, it is called a hexaedron; if eight, an octaedron • if twelve, a dodecaedron; if twenty, an icosaedron, &c.

2. The intersections of the faces of a polyedron are called 'ts edges. A diagonal of a polyedron is the straight line which joins any two vertices not lying in the same face.

3. Similar polyedrons are such as have all their solid angles equal, each to each, and are contained by the same number of similar polygons.

4. A regular polyedron is one whose solid angles are all equal to each other, and whose faces are all equal and regu lar polygons.

5. A prism is a polyedron having two faces which are equal and parallel polygons; and the others are parallelograms. The equal and parallel polygons are called the bases of the prism; the other faces taken together form the lateral or convex surface. The altitude of a prism is the perpendicular distance between its two bases. The edges which join the corresponding angles of the two polygons are called the principal edges of the prism.

6. A right prism is one whose principal edges are all per pendicular to the bases. Any other prism is called an ob lique prism

7. A prism is triangular, quadrangular, pentagonal, hexagonal, &c., according as its base is a triangle, a quadrilateral, a pentagon, a hexagon, &c.

8. A parallelopiped is a prism whose bases are parallelograms. A right parallelopiped is one whose faces are all rectangles.

9. A cube is a right parallelopiped bounded by six equa.

squares,

10. A pyramid is a polyedron contained by several triangular planes proceeding from the same point, and terminating in the sides of a polygon. This polygon is called the base of the pyramid; and the point in which the planes meet, is the vertex. The triangular planes form the convex surface.

11. The altitude of a pyramid is the perpendicular let fall from the vertex upon the plane

of the base, produced if necessary. The slant height of a pyramid is a line drawn from the vertex, perpendicular to one side of the polygon which forms its base.

12. A pyramid is triangular, quadrangular, &c., according as the base is a triangle, a quadrilateral, &c.

13. A regular pyramid is one whose base is a regular polygon, and the perpendicular let fall from the vertex upon the base, passes through the center of the base. This perpendic ular is called the axis of the pyramid.

14. A frustum of a pyramid is a portion of the solid next the base, cut off by a plane parallel to the base. The altitude of the frustum is the perpendicular distance between the two parallel planes.

PROPOSITION I. THEOREM.

The convex surface of a right prism is equal to the perimeter of its base multiplied by its altitude.

Let ABCDE-K be a right prism; then will its convex surface be equal to the perimeter of the base of AB+BC+CD+DE+EA multiplied by its altitude AF.

H

C

For the convex surface of the prism is equal to the sum of the parallelograms AG, BH, CI, &c. Now the area of the parallelo- A gram AG is measured by the product of its base AB by its altitude AF (Prop. IV., Sch., B. IV.). The area of the parallelogram BH is measured by BCXBG; the area of CI is measured by CDXCH, and so of the others. But the lines AF, BG, CH, &c., are all equal to each other (Prop. XIV., B. VII.), and each equal to the altitude of the prism. Also, the lines AB, BC, CD, &c., taken together, from the perimeter of the base of the prism. Therefore, the sum of these parallelograms, or the convex surface of the prism, is equal to the perimeter of its base, multiplied by its altitude.