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therefore in the triangles FHC, FKC, there are two angles of the one equal to two angles of the other, each to each;

and the side FC, which is opposite to one of the equal angles in each, is common to both;

therefore the other sides are equal, each to each; (I. 26.) wherefore the perpendicular FH is equal to the perpendicular FK: in the same manner it may be demonstrated, that FL, FM, FG are each of them equal to FH, or FK:

therefore the five straight lines FG, FH, FK, FL, FM are equal to one another:

wherefore the circle described from the center F, at the distance of one of these five, will pass through the extremities of the other four, and touch the straight lines AB, BC, CD, DE, EA,

because the angles at the points G, H, K, L, M are right angles, and that a straight line drawn from the extremity of the diameter of a circle at right angles to it, touches the circle; (III. 16.) therefore each of the straight lines AB, BC, CD, DE, EA touches

the circle:

wherefore it is inscribed in the pentagon ABCDE. Q.E.F.

PROPOSITION XIV. PROBLEM.

To describe a circle about a given equilateral and equiangular pentagon. Let ABCDE be the given equilateral and equiangular pentagon. It is required to describe a circle about ABCDE.

A

B

C

D

Bisect the angles BCD, CDE by the straight lines CF, FD, (1. 9.) and from the point F, in which they meet, draw the straight lines FB, FA, FE, to the points B, A, E.

It may be demonstrated, in the same manner as the preceding proposition,"

that the angles CBA, BAE, AED are bisected by the straight lines FB, FA, FE.

And because the angle BCD is equal to the angle CDE,
and that FCD is the half of the angle BCD,
and CDF the half of CDE;

therefore the angle FCD is equal to FDC; (ax. 7.) wherefore the side CF is equal to the side FD: (1.6.) in like manner it may be demonstrated, that FB, FA, FE, are each of them equal to FC or FD: therefore the five straight lines FA, FB, FC, FD, FE, are equal to

one another;

and the circle described from the center F, at the distance of one of them, will pass through the extremities of the other four, and be described about the equilateral and equiangular pentagon ABCDE.

Q.E.F.

PROPOSITION XV. PROBLEM.

To inscribe an equilateral and equiangular hexagon in a given circle. Let ABCDEF be the given circle.

It is required to inscribe an equilateral and equiangular hexagon in it.

A

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H

Find the center G of the circle ABCDEF,
and draw the diameter AGD; (III. 1.)

and from D, as a center, at the distance DG, describe the circle EGCH,
join EG, CG, and produce them to the points B, F;
and join AB, BC, CD, DE, EF, FA:

the hexagon ABCDEF shall be equilateral and equiangular.
Because G is the center of the circle ABCDEF,
GE is equal to GD:

and because D is the center of the circle EG CH,
DE is equal to DG:

wherefore GE is equal to ED, (ax. 1.)
and the triangle EGD is equilateral;

and therefore its three angles EGD, GDE, DEG, are equal to one another: (I. 5. Cor.)

but the three angles of a triangle are equal to two right angles; (1. 32.) therefore the angle EGD is the third part of two right angles: in the same manner it may be demonstrated,

that the angle DGC is also the third part of two right angles: and because the straight line GC makes with EB the adjacent angles

EGC, CGB equal to two right angles; (1. 13.)

the remaining angle CGB is the third part of two right angles: therefore the angles EGD, DGC, CGB are equal to one another: and to these are equal the vertical opposite angles BGA, AGF, FGE:

(I. 15.)

therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE, are equal to one another :

but equal angles stand upon equal circumferences; (III. 26.) therefore the six circumferences AB, BC, CD, DE, EF, FA are equal

to one another:

and equal circumferences are subtended by equal straight lines: (III. 29.)

therefore the six straight lines are equal to one another,
and the hexagon ABCDEF is equilateral.
It is also equiangular:

for, since the circumference AF is equal to ED,

to each of these equals add the circumference ABCD; therefore the whole circumference FABCD is equal to the whole EDCBA:

and the angle FED stands upon the circumference FABCD,
and the angle AFE upon EDCBA;

therefore the angle AFE is equal to FED: (III. 27.)
in the same manner it may be demonstrated,

that the other angles of the hexagon ABCDEF are each of them equal to the angle AFE or FED: therefore the hexagon is equi

angular; and it is equilateral, as was shewn;

and it is inscribed in the given circle ABCDEF. Q.E.F. COR.-From this it is manifest, that the side of the hexagon is equal to the straight line from the center, that is, to the semi-diameter of the circle.

And if through the points A, B, C, D, E, F there be drawn straight lines touching the circle, an equilateral and equiangular hexagon will be described about it, which may be demonstrated from what has been said of the pentagon: and likewise a circle may be inscribed in a given equilateral and equiangular hexagon, and circumscribed about it, by a method like to that used for the pentagon.

PROPOSITION XVI.

PROBLEM.

To inscribe an equilateral and equiangular quindecagon in a given circle. Let ABCD be the given circle.

It is required to inscribe an equilateral and equiangular quindecagon in the circle ABCD.

B

E

A

D

F

Let A Cbe the side of an equilateral triangle inscribed in the circle, (IV.2.) and AB the side of an equilateral and equiangular pentagon inscribed

in the same; (IV. 11.)

therefore, of such equal parts as the whole circumference ABCDF

contains fifteen,

the circumference ABC, being the third part of the whole, contains five; and the circumference AB, which is the fifth part of the whole, contains three;

therefore BC, their difference, contains two of the same parts: bisect BC in E; (III. 30.)

therefore BE, EC are, each of them, the fifteenth part of the whole

circumference ABCD:

therefore if the straight lines BE, EC be drawn, and straight lines equal to them be placed round in the whole circle, (IV. 1.) an equilateral and equiangular quindecagon will be inscribed in it. Q. E. F.

And in the same manner as was done in the pentagon, if through the points of division made by inscribing the quindecagon, straight lines be drawn touching the circle, an equilateral and equiangular quindecagon will be described about it: and likewise, as in the pentagon, a circle may be inscribed in a given equilateral and equiangular quindecagon, and circumscribed about it.

THE Fourth Book of the Elements contains some particular cases of four general problems on the inscription and the circumscription of triangles and regular figures in and about circles. Euclid has not given any instance of the inscription or circumscription of rectilineal figures in and about other rectilineal figures.

Any rectilineal figure, of five sides and angles, is called a pentagon; of seven sides and angles, a heptagon; of eight sides and angles, an octagon; of nine sides and angles, a nonagon; of ten sides and angles, a decagon; of eleven sides and angles, an undecagon; of twelve sides and angles, a duodecagon; of fifteen sides and angles, a quindecagon, &c.

These figures are included under the general name of polygons; and are called equilateral, when their sides are equal; and equiangular, when their angles are equal; also when both their sides and angles are equal, they are called regular polygons.

Prop. 11. An objection has been raised to the construction of this problem. It is said that in this and other instances of a similar kind, the lines which touch the circle at A, B, and C, should be proved to meet one another. This may be done by joining AB, and then since the angles KĀM, KBM are equal to two right angles (111. 18.), therefore the angles BAM, ABM are less than two right angles, and consequently (ax. 12.), AM and BM must meet one another, when produced far enough. Similarly, it may be shewn that AL and CL, as also CN and BN meet one another.

Prop. v. is the same as "To describe a circle passing through three given points, provided that they are not in the same straight line."

The corollary to this proposition appears to have been already demonstrated in Prop. 31. Book III.

It is obvious that the square described about a circle is equal to double the square inscribed in the same circle. Also that the circumscribed square is equal to the square on the diameter, or four times the square on the radius of the circle.

Prop. vii. It is manifest that a square is the only right-angled parallelogram which can be circumscribed about a circle, but that both a rectangle and a square may be inscribed in a circle.

Prop. x. By means of this proposition, a right angle may be divided into five equal parts.

Reference has already been made to the distinction between analysis and synthesis, and that all Euclid's direct demonstrations are synthetic, properly so called. There is however a single exception in Prop. 16. Book Iv, where the analysis only is given of the Problem. The two methods are so connected in all processes of reasoning, that it is very difficult to separate one from the other, and to assert that this process is really synthetic, and that is really analytic. In every operation performed in the construction of a problem, there must be in the mind a knowledge of some properties of the figure which suggest the steps to be taken in the construction of it. Let any Problem be selected from Euclid, and at each step of the operation, let the question be asked, “Why that step is taken?" It will be found that it is because of some known property of the required figure. As an example will make the subject more clear to the learner, the Analysis of Euc. iv. 10, is taken from the "Analysis of Problems" in the larger edition of the Euclid, and to which the learner referred for more complete information.

Euc. Iv. 10, there are five operations specified in the construc

Take any straight line AB.

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(2) Divide the line AB in C, so that the rectangle AB, BC, may be equal to the square on AC.

(3) Describe the circle BDE with center A and radius AB. (4) Place the line BD in that circle, equal to the line AC. (5) Join the points A, D.

Why should either of these operations be performed rather than any others? And what will enable us to forsee that the result of them will be such a triangle as was required? The demonstration affixed to it by Euclid does undoubtedly prove that these operations must, in conjunction, produce such a triangle: but we are furnished in the Elements with no obvious reason for the adoption of these steps, unless we suppose them accidental. To suppose that all the constructions, even the simpler ones, are the result of accident only, would be supposing more than could be shewn to be admissible. No construction of the problem could have been devised without a previous knowledge of some of the properties of the figure. In fact, in directing the figure to be constructed, we assume the possibility of its existence; and we study the properties of such a figure on the hypothesis of its actual existence. It is this study of the properties of the figure that constitutes the Analysis of the problem.

Let then the existence of a triangle BAD be admitted, which has each of the angles ABD, ADB double of the angle BAD, in order to ascertain any properties it may possess which would assist in the construction of such a triangle.

Then, since the angle ADB is double of BAD, if we draw a line DC to bisect ADB and meet AB in C, the angle ADC will be equal to CAD; and hence (Euc. 1. 6.) the sides AC, CD are equal to one another.

Again, since we have three points A, C, D, not in the same straight line, let us examine the effect of describing a circle through them: that is, describe the circle ACD about the triangle ACD. (Euc. Iv. 5.)

Then, since the angle ADB has been bisected by DC, and since ADB is double of DAB, the angle CDB is equal to the angle DAC in the alternate segment of the circle; the line BD therefore coincides with a tangent to the circle at D. (Converse of Euc. III. 32.)

Whence it follows, that the rectangle contained by AB, BC, is equal to the square on BD. (Euc. III. 36.)

But the angle BCD is equal to the two interior opposite angles CAD, CDA; or since these are equal to each another, BCD is the double of CAD, that is, of BAD. And since ABD is also double of BAD, by the conditions of the triangle, the angles BCD, CBD are equal, and BD is equal to DC, that is, to AC.

It has been proved that the rectangle AB, BC, is equal to the square on BD; and hence the point C in AB, found by the intersection of the bisecting line DC, is such, that the rectangle AB, BC is equal to the square on AC. (Euc. 11. 11.)

Finally, since the triangle ABD is isosceles, having each of the angles ABD, ADB double of the same angle, the sides AB, AD are equal, and hence the points B, D, are in the circumference of the circle described about A with the radius AB. And since the magnitude of the triangle is not specified, the line AB may be of any length whatever.

From this "Analysis of the problem," which obviously is nothing more than an examination of the properties of such a figure supposed to exist already, it will be at once apparent, why, those steps which are prescribed by Euclid for its construction, were adopted.

The line AB is taken of any length, because the problem does not prescribe any specific magnitude to any of the sides of the triangle.

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