Join GA, GB, GC, and DE, EF, FD. Then, putting a, b, c to denote the sides opposite to the angles A, B, C, we have (VI. 2. and 4.) FE=a, FD=b, and DE

B

D

E

c; and (APP. I.4.) since AEG, AFG are right angles, a circle may be described about the quadrilateral AEGF. For a like reason circles may be described about BDGF and CDGE. Hence (VI. E.) FE.AG = AF.GE+ AE.FG; or, by doubling, a.AG = c.GE+b.FG. In the same manner it would be shown, since AG, BG, CG are equal, that b.AG = c.GD + a.FG, and c.AG= a.GE+b.GD. Hence, by addition, (a+b+c) AG=(a+c) GE +(a+b) GF+ (b+c) GD. Now b.GE is evidently equal to twice the triangle AGC, c.GF equal to twice AGB, and a.GD equal to twice BGC: also, denoting the radius of the inscribed circle by r, we have (IV. 4. cor. 6.) r(a + b + c) equal to twice the area of the triangle ABC, and consequently r(a+b+c)= b.GE+c.GF+ a.GD. Hence, by addition, (a+b+c) AG + r(a+b+c)=(a+b+c) GE+(a+b+c) GF+(a+b+c) GD; and consequently AG + GE+GF+ GD.

Cor. Since, by the scholium to proposition XLIII. of this book, the parts of the three perpendiculars of the triangle, between their common intersection and the three angles, are respectively double of GD, GE, GF, the sum of those parts of the perpendiculars is equal to the sum of the diameters of the inscribed and circumscribed circles.

BOOK V.

PROP. I. PROB.

To draw a straight line perpendicular to two given straight lines, not in the same plane.

Let AB, CD be two given straight lines not in the same plane; it is required to draw a straight line which will be perpendicular to them both.

Through any point D in one of them, draw (I. 31.) DE parallel to the other, and (XI. 12.) through the same point D draw DF perpendicular to the plane passing through DE, DC: let the plane passing through CD, DF meet AB in A, and (I. 12.) from A draw AC perpendicular to CD; AC is the perpendicular required.

Draw AF parallel to CD. Then (I. 28.) CA, DF are parallel, because they are both perpendicular to CD. FA, AB, therefore, being parallel to CD, DE, the plane passing through FA, AB is parallel (XI. 15.) to the one passing through CD, DE; and therefore FD, being (const.) perpendicular to the plane through CD,

F

D

E

B

DE, is also, by the demonstration of the 15th proposition of the eleventh book, perpendicular to the one through FA, AB; and therefore AC, which is parallel to FD, is also perpendicular (XI. 8.) to the plane through FA, AB, and consequently (XI. def. 1.) to the straight line AB: and being (const.) perpendicular to CD, it is the line required.

Schol. Through the two straight lines AB, CD, which are not parallel, two parallel planes can be drawn-those passing through AB, AF, and through CD, DE: but no other parallel planes

besides these can be drawn through AB, CD, as AB can meet the plane CF in only one point. For the same reason, there is only one common perpendicular to AB, CD. On the contrary, were AB, CD parallel, there might be innumerable parallel planes, and common perpendiculars. It is also evident, since AC is the perpendicular distance between the two planes, that it is less than any line between them which is not perpendicular to both, and therefore that it is the least straight line that can be drawn between AB and CD.

PROP. II. THEOR.

THE sections of a prism by parallel planes, are equal and similar figures.

M

Let the prism MN be cut by the two parallel planes ABCDE, FGHKL; their sections with it are equal and similar figures. For (XI. 16.) AB, FG are parallel, and (I. 34.) they are equal. For the same reason BC, GH are equal and parallel; as are also CD, HK; DE, KL; and AE, FL. The figures AD, FK, therefore, have their sides equal, each to each. Their several angles are also (XI. 10.) equal; for they are contained by straight lines which are parallel: F and therefore the figures are equal and similar.

Cor. A section of a prism by a plane parallel to the base, is equal and similar to the base.

A

E

D

K

G

H

N

Schol. Since (XI. def. 24.) a cylinder is described by the revolution of a rectangle about one of its sides, it is plain that any straight line in the rectangle perpendicular to the fixed line, will describe a circle parallel to the base; and hence every section of a cylinder by a plane parallel to the base, is a circle equal to the base.

PROP. III. THEOR.

THE section of a pyramid by a plane parallel to its base, is a figure similar to the base.

Let the pyramid ABCDE be cut by the plane FGHK parallel to its base BCDE; the section FGHK is similar to the base.

It would be proved as in the preceding proposition, that the figures GK, CE are equiangular. Then, because the triangles

Hence

AFG, ABC are equiangular, FG: BC:: AG: AC; and because AGH, ACD are equiangular, GH: CD:: AG : AC. (V. 11. and alternately) FG: GH:: BC: CD; and in a similar manner it would be proved, that the sides about the other angles of GK, CE are proportionals. (VI. def. 1.) GK is similar to CE.

Hence

G

K

M

H

B

E

Cor. Draw AML perpendicular to CE, and consequently to GK: the altitude AL, and the straight lines AB, AC, AD, AE are cut proportionally by GK. Join CL, GM: these (XI. 2. and 16.) are in the same plane, and are parallel to one another. Hence (VI. 2.) AF: FB:: AG: GC: AM: ML, &c.

Schol. 1. Since (XI. def. 21.) a cone is described by the revolution of a right-angled triangle about one of its legs, it is plain that any straight line in the triangle perpendicular to the fixed leg, will describe a circle parallel to the base; and the radius of that circle will be to the radius of the base, as the altitude of the cone cut off to that of the whole cone.

Schol. 2. The pyramid or cone cut off from another pyramid or cone, by a plane parallel to the base, is similar (XI. def. 8. and 27.) to the whole pyramid or cone.

PROP. IV. THEOR.

If the altitude of a parallelepiped, and the length and perpendicular breadth of its base be all divided into parts equal to one another, the continued product of the number of parts in the three lines is the number of cubes contained in the parallelepiped, each cube having the side of its base equal to one of the parts.

First, suppose the parallelepiped to be rectangular. Then planes parallel to the base passing through the points of section of the altitude, will evidently divide the solid into as many equal solids, as there are parts in the altitude; and each of these partial solids will be composed of as many cubes, as the base contains squares each equal to a base of one of the cubes. But (I. 46. cor. 4.) the number of these squares is the product of the length and breadth of the base: and hence the entire number of cubes will be equal to the product of the three dimensions, the length, breadth, and altitude.

If the base be not rectangular, its area (I. 46. cor. 5.) will be the product of its length and perpendicular breadth and it is evident that the product of this by the altitude, will be the number of cubes as before.

Lastly, if the insisting lines be not perpendicular to the base, still the oblique parallelepiped is equal (XI. 29. and 30.) to a rectangular one of the same altitude; and therefore the number of cubes will be found as before, by multiplying the area of the base by the altitude.

Cor. 1. Hence it is evident that the volume, or numerical solid content, or, as it is also called, the solidity, of a parallelepiped, is the product of its altitude, and the area of its base, both expressed in numbers and it is plain, that the same holds in regard to any prism whatever, and also in regard to cylinders.*

Cor. 2. The content of a pyramid or cone is found by multiplying the area of the base by the altitude, and taking a third of the product. For (XII. 7. cor. 1.) a pyramid is a third part of a prism, and (XII. 10.) a cone a third part of a cylinder, of the same base and altitude.

Cor. 3. The content of a polyhedron may be found by dividing it into pyramids, and adding together their contents. The division into pyramids may be made either by planes passing through the vertex of one of the solid angles, or by planes passing through a point within the body.

PROP. V. THEOR.

THE surfaces of two similar polyhedrons may be divided into the same number of similar triangles, similarly situated.

This follows immediately from the definition (XI. def. 8.) of similar bodies bounded by planes, if the sides or faces of the polyhedron be triangles: and any face in the one, and the corresponding face in the other, which are not triangles, are yet similar, and may be divided (VI. 20.) into the same number of similar triangles, similarly situated.

Cor. Hence it would be shown, as in the 20th proposition of the sixth book, that the surfaces of the polyhedrons are proportional to any two of their similar triangles; and therefore (VI. 19.) that they are to one another in the duplicate ratio of the homologous sides of those triangles, that is, of the edges or intersections of the similar planes. Hence also the surfaces are proportional (VI. 20. cor. 3.) to the squares of the edges.

As an example, suppose the length, breadth, and depth of a rectangular chest to be respectively 4 feet, 3 feet, and 2 feet: its content will be 4 X 3X2 = 24 cubic feet; that is, 24 cubes having their bases each a square foot. The measuring unit is here a cubic foot; but it is evident, that it may, according to convenience, be a cubic inch, a cubic yard, &c.

As another example, let the radius of the base of a cylinder be 10 inches, and its altitude 15 inches. Then (APP. I. 39. cor. 2.) the area of the base is 314-1593; the product of which by 15 is 4712-3895 cubic inches, the content of the cylinder.