tended arcs into two, four, &c. equal parts (54.) a regular figure of twice, four times, &c. the number of sides of any of the above may be inscribed; e.g. an octagon, (or regular figure of eight sides,) by bisecting the arcs which are subtended by the sides of a square, a dodecagon (or regular figure of twelve sides) by bisecting those which are subtended by the sides of a hexagon; and so on. Still there are many regular figures, as the heptagon, enneagon, hendecagon, &c. (figures of 7, 9, 11, &c. sides), for the inscribing of which no exact geometrical rule has ever been discovered. These figures we can only inscribe approximatively; and when it is required to do so with a considerable degree of accuracy, the following method may be adopted.

Let it be required, for instance, to inscribe a regular heptagon. Continue the series 4, 8, 16, &c. which represents the numbers of parts into which the circumference may be divided by continued bisections, until a number be found which is greater or less by 1 than a multiple of 7: 64 is such a number, being greater by 1 than 9 x 7.

Now, if the circumference be divided into 64 equal parts, and the arc AG be taken equal to 9 of those parts (which may be done by bisecting the quadrantal arc AB in D, DB in E, DE in F, and D F in G), the arc A G will be less than a seventh part of the circumference by a seventh part of one of them DG. But, the arc DG being small, a seventh part of its chord (which may be found by (I. 49.) may without any considerable error be assumed for the seventh part of the arc itself, being somewhat less than the latter; and if the chord of A a be taken equal to this approximate seventh part, the error of assuming for it the arc Aa, which is somewhat greater than its chord, will be still less, so that G a will be equal, very nearly, to one-seventh of the circumference, and the chord of G a very nearly equal to the side of a regular heptagon inscribed in the circle.

For a second example, let us take the enneagon, or, as it is sometimes called, nonagon: here, again, 64 exceeds 7 × 9 by 1; therefore, the same division of the quadrant being made as in the case of the heptagon, because B G is equal to 7 parts out of the 64, and DG to one part, B G with a ninth of D G will be contained in the circumference 9 times exactly and if the chord of Bb be taken equal to the ninth part of the chord of DG, the chord of Gb will be very nearly equal to the side of a regular enneagon inscribed in the circle.

It is not, however, necessary that we should always proceed with the series till we arrive at a number greater or less than the number of sides by 1. Take, for instance, the hendecagon, or, as it is sometimes called, undecagon: here 11 x 6

66, which exceeds 64 by 2. Now the arc B F contains 6 out of 64 parts DG of the circumference. Therefore, if the circumference be increased by twice DG, BF will be contained in the circumference so increased 11 times, and, consequently, if B F be diminished by twoelevenths of DG, it will be contained in the circumference 11 times exactly; so that the side of the hendecagon will be obtained approximatively by assuming, as before, the chord in place of the arc, and taking from B F two-elevenths of the former instead of two-elevenths of the latter.

In these examples, the real errors, if computed, will be found far more minute than those which the imperfection of our instruments entails upon the most accurate geometrical constructions. Seven times the arc which has been assumed as a seventh of the circumference falls short of the whole circumference by less than the Tooth part, and 9 times the arc which has been assumed as a ninth, by about an equal quantity; while 11 times the arc, which has been assumed as an eleventh, exceeds by only about twice the same quantity. The method may therefore be adopted in these and in similar cases, as practically accurate.

PROP. 64. Prob. 11.

To construct a triangle, any three being assumed out of the four following data, viz. the vertical angle, the base, the sum of the sides, and the area.

This problem comprehends four cases, in which the data are respectively, 1. Vertical angle, base, and sum of sides;

2. Vertical angle, base, and area;

1. Let AB

be the given base, AC the given sum of the two sides, and Ꭰ the given vertical angle. Upon

F

E

D

A B (60.) describe a segment AEB, containing an angle equal to the half of D. With the centre A and radius A C describe a circle cutting the arc A E B in E. Join A E, B E, and at the point B make the angle EBF equal to BEA: the triangle F Ä B shall be the triangle required. For, because the angle FEB is equal to FB E, the side F B is equal to FE (I. 6.), and the two sides AF, FB together are equal to AE, that is, to AC, the given sum of the sides. Again, because the angle AFB (I. 19.) is equal to the sum of the angles at E and B, and that these angles are equal to one another, the angle AFB is equal to twice the angle at E, that is, to the given angle D. And the triangle is described upon the given base A B. Therefore, &c.

2. Let A B be the given base, upon which let there be described the rectangle A B CD, contain

DE

ing an area equal to twice the given area (I. 57.), and a segment AEB containing an angle equal to the given angle (60.). Then if the are A E B cut the side CD in E, and EA, EB be joined, E A B will evidently be the triangle required.

area

E

B

3. Let A be the given vertical angle, and let the triangle ABC (II.69.) contain an equal to the given area: and let D be the given sum of the two sides. Divide D into two parts, such that their rectangle may be equal to the rectangle under AB, AC (II.56.). Take AE equal to one, and AF equal to the other of these parts, and join EF. Then, because the triangles ABC, AEF have the common angle A, they are to one another (II.40. Cor.) as the rectangles under the con

D draw DG parallel to AB (I. 48.) to meet FG in G, and join G A. Through E and C draw EH and C K parallel to FG, to meet GA and GA produced in the points H, K, so that K G, KH, KA will be proportionals (II. 29.): from the centre K, with the radius KH, describe a circle cutting GD in L, and join LA, LB: LAB shall be the triangle required.

Produce G K to M, so that K M may be equal to KH; and from L draw LŇ perpendicular to AB. Then, because KÁ, KH, KG are proportionals, MA, MH, MG are in harmonical progression (II. 46.); but the point L is in the circumference of a circle upon the mean MH; therefore (51. Cor.) LA: LG:: AH: HG; but LG is (I. 22.) equal to NF, and AH: HG :: AE EF (II. 29.), that is, since CA, CE, CF are proportionals (II. 22. Cor. 1.)::CA: CE; therefore LA is to N F as CA to CE, or in the subduplicate ratio of C A to CF; and (II. 38. Schol. Lem. 1. Cor.) if Bf be taken equal to A F, LB is to Nf in the same ratio. Therefore the sum of LA, LB is to the whole line Ff in the same ratio, or, if Ce be taken equal to CE, as Ee to Ff: therefore, the sum of LA, LB is equal (II. 11. Cor. 1.) to E e, that is, to the given sum, and LAB is the triangle required. Therefore, &c.*

If the difference of the sides be supposed given instead of the sum in cases 1, 3 and 4, solutions of the same character may be obtained; viz. in case by describing upon the given base AB a segment which shall contain an angle exceeding by a right angle half the given angle D; in case 3, by dividing the difference D produced, so that the rectangle under the segments may be equal to A BXA C; and in case 4, by making use of the following corollary to

PROP. 65. Prob. 12.

To find two straight lines, there being assumed any two out of the six following data; viz. their sum, their difference, the sum of their squares, the dif ference of their squares, their ratio, and their rectangle.

The cases of this proposition are fifteen in number, and may be arranged as follows:

1. Sum, and difference.

2. Sum of squares, and difference of squares.

3. Sum, and sum of squares.
4. Difference, and sum of squares.
5. Sum, and difference of squares.
6. Difference, and difference of squares.

7. Ratio, and rectangle (II. 63.) 8. Sum, and ratio (II. 55. fig. 1.) 9. Difference, and ratio (II. 55. fig. 2.) 10. Sum, and rectangle (II. 56. fig. 1.) 11. Difference, and rectangle (II. 56. fig. 2.)

12. Sum of squares, and ratio.

13. Difference of squares, and ratio. 14. Sum of squares, and rectangle. 15. Difference of squares, and rectangle.

Those of the second division, viz. the 7th, 8th, 9th, 10th, and 11th cases, have in effect been already considered in the propositions referred to at the side. They will accordingly be here omitted. Of the rest the greater part are so ob. vious, that it will be sufficient to indicate only the construction, leaving the

demonstration to the reader.

1. Let AB be the given sum, AC the given difference. Bisect CB in D, and

and CD in F, and take E G a third proportional to EB and C F. From the centre E, with the radius EA, describe a semicircle AHB; and from the point G (I. 44.) draw GH perpendicular to A B, to meet the circumference in H. Join A H, HB: they shall be the straight lines required.

I

3. Let AB be the given sum, and the square of AC the given sum of the squares. Bisect AC in D, (I. 43.) and from D (I.44.) draw DE perpendicular to A C. From the centre D

B

circle cutting DE in E; and from the with the radius DA or DC describe a centre E, with the radius EA or EC describe the circle C FA: lastly, from the centre A with the radius A B describe an arc cutting C F A in F. Join AF, let AF cut the circumference AEC in G: and join G C. AG and GC shall be the lines required.

4. Let A B be the given difference, and the square of A C the given sum of the squares. Bisect AC in D (I. 43.), and from D (I. 44.) draw DE perpendicular to AC: from the centre

D with the radius

DA or DC describe in E; and from the a circle cutting DE

E

G

B

centre E with the radius EA or EC de scribe the circle CFA; lastly, from the centre A with the radius AB describe let AF produced cut the circumference an arc cutting CFA in F: join AF, and AECG in G, and join GC. AG and GC shall be the straight lines required.

5. Let A B be the given sum, and the square of A C the given difference of the squares: take AD a third pro

and the square of A C the given difference of the squares: take A D a third proportional to A B, A C (II. 52.), and

would be impossible. In this manner does one of the conditions frequently set limits to the other-frequently, but not in every case:-thus, if two lines be required, which shall contain a given rectangle, their ratio may be any whatever, and a problem which should require them to be to one another in any ratio, how great or how small soever, would be possible. The limits of possibility, when there are any, are commonly indicated by the construction, if the problem be solved geometrically, as they are, if algebraically, by the form of the final equation. See the cases of Prop. 64., where the vertex of the triangle sought is determined by the intersection of a straight line and circle, or of two circles: if the data be such that no intersection can take place, the construction fails, and the problem becomes impossible.

BOOK IV.

§1. Of Lines perpendicular, or inclined, or parallel to planes.-§ 2. Of Planes which are parallel, or inclined, or perpendicular to other Planes.-§ 3. Of Solids contained by Planes. § 4. Problems.

SECTION 1.—Of Lines perpendicular, or inclined, or parallel to Planes. IN the preceding books our attention has been confined to lines which lie in one and the same plane, the intersection of such lines, and the figures contained by them; we are now to consider lines which lie in different planes, planes which intersect one another, and solids which are contained by plane or other surfaces. In other words, we have been hitherto engaged with Plane Geometry; we are now to enter upon Solid Geometry.

Def. 1. (Euc. xi. def. 3.) A straight line is said to be perpendicular (or at right angles) to a plane, when it makes right angles with every straight line meeting it in that plane, (see Prop 3.). Also, conversely, in this case the plane is said to be perpendicular to the straight line.

The foot of the perpendicular is the point in which it meets the plane.

It is evident that a straight line cannot meet a plane in more than one point, unless it lies altogether in the plane; and in like manner that one plane cannot meet another plane in a portion of sur

2. (Euc. xi. def. 5.) A straight line is said to be inclined to a plane, when it meets the plane but is not perpendicular to it.

C

When a straight line A B is inclined to a plane CDĚ, the angle ABF which it makes with a straight line drawn from the point B in which it meets the plane,through D the foot of the

B

F

perpendicular AF, which is let fall upon the plane from any other point A of the straight line (see Prop. 7.), is called the angle of inclination.

3. A straight line is said to be parallel to a plane, when it cannot meet the plane, to whatever extent both be produced. Also, conversely, in this case the plane is said to be parallel to the straight line.

4. If two planes ABC, ABD intersect one another in a line as A B (see Prop. 2.), they are

said to form at that line

a dihedral angle CABD.

A

B

D

The magnitude of a dihedral angle does not depend upon the extent of the containing planes, but upon the opening between them. Thus, the dihedral angle CABD is greater than dral angle CABE. the dihedral angle EA BD by the dihe

another plane makes the adjacent dihe5. When one plane standing upon dral angles equal to one another, each

of them is called a right dihedral angle; and the plane which stands the other upon

is said to be perpendicular (or at right angles) to it.

A dihedral angle is also said to be acute or obtuse, according as it less or greater than a right angle.

6. (Euc. xi. def. 8.) Planes, which do not meet one another, though produced to any extent, are said to be parallel.

7. (Euc. xi. def. 9.) If three or more planes pass through a point as A, they are

face common to both, unless they coincide altogether. (See Prop. 1.) Therefore a straight line cuts a plane in a point; and a plane cuts a plane in a line, which line (see Prop. 2.) is a straight line,