one another; that is, it is bisected by

the line A D.

And, in the same words, it may be demonstrated, that if the exterior vertical angle be bisected by the line A d, which cuts the base produced in d, Bd shall be to dC as BA to AC; and that, conversely, if Bd be to d C as BA to AC, Ad shall bisect the exterior vertical angle: the letters d, e, being substituted for D, E. Therefore, &c.

When the sides AB, AC are equal to one another, A D bisects the base B C at right angles (I. 6. Cor. 3.); and Ad is parallel to the base BC; for Ad is always at right angles to AD, because the angle DA d is equal to the halves of the two angles BAC, CAE together, that is, to the half of two right angles.

Cor. Since the ratios of B D to DC, and of B d to d C, are each of them the same with the ratio of B A to A C, they

are the same with one another (12.), and

Bd is harmonically divided in the points D, C. Therefore the two sides of a triangle, and the lines which bisect the vertical and exterior-vertical angles, are harmonicals (def. 20.).

SECTION 7.-Problems.

Def. 21. A straight line is said to be divided in extreme and mean ratio, when

the whole line is to the greater segment as the greater segment is to the less.

A straight line so divided is also said to be divided medially; and the ratio of its segments is called the medial ratio.

PROP. 51. Prob. 1. (Euc. vi. 13.) To find a mean proportional between two given straight lines A B and B C. Let the straight lines A B, BC be placed in the same straight line: from the point B (I. 44.) draw B D at right angles to AC: bisect A C in E (I. 43.), and from the centre E, with the radius E C, describe a circle cutting BD in D: BD shall be the mean proportional required.

For, the angle EBD being a right angle, the square of B D is (1.36.Cor. 1.) equal to the difference of the squares of EB, ED. But, because E A and E C are, each of them, equal to ED, AB is equal to the sum, and B C to the difference of ED, E B: therefore (1.34.) the square of BD is equal to the rectangle under AB, BC, and (38. Cor. 1.) BD

(I. 48.), to meet AE produced in F.

D

Then, because (29.) A B is to B C as proportional required. AE (or D) to EF, EF is the fourth Therefore, &c.

PROP. 54. Prob. 4. (Euc. vi. 10.) To divide a given straight line A, similarly to a given divided straight line BC. Method 1. Draw BD making any angle with B C, and make BD equal to A: join C D, and through the several points in which BC is divided, draw lines parallel to CD (I.48.): then

A

(29. Cor.), because these lines are parallel to CD, the straight line DB, that is, A, is divided by them similarly to the given divided straight line B C. Method 2. Upon BC describe (I.42.) the equilateral triangle DB C: take DE, DF each of them equal to A, and join EF; and from D through the several points in which B C

divided, draw straight lines cutting EF.

E

Then, because DE is equal to D F, and DB to DC, the triangles DEF, DBC are similar (32.): but BC is equal to BD; therefore E F is equal to ED, that is, to the given straight line A; and, for the same reason, the angle DEF is equal to the angle DBC: therefore (I. 15.) E F is parallel to BC. And, because EF, BC are parallels, EF, that is, A, is divided similarly to BC by lines drawn from the point D (30. Cor. 1.).

squares of D B, DE (I. 34.), that is, to the square of C; and AB is divided, as required, in the point E. Therefore, &c.

Cor. Hence also a given straight line AB or AB produced may be so divided, that the rectangle under the segments may be equal to a given rectangle (I. 58.).

N. B. In the first case, viz. when the point E is to be found between A and B, the problem will be impossible if C be greater than the half of AB; for AE × EB is equal to the difference of the squares of DB, DE, which is never greater than the square of DB. The second case is not thus limited.

We may remark also, that in both cases, two points E may be found satisfying the given conditions, viz. one upon each side of the point D: in the first case, when C is equal to the half of AB, these two points coincide at D.

PROP. 57. Prob. 7.

A

D B

To find an harmonical mean between two given straight lines A B and A C. Divide BC in the point Din the ratio of B A to A C (55.): then, because AB: AC::BD: DC, the three lines AB, AD, and AC are in harmonical progression (def. 17.), and AD is an harmonical mean between A B and AC Therefore, &c.

PROP. 58. Prob. 8.

To find a third harmonical progressional to two given straight lines A B, A C.

Divide AC produced in the point D in the ratio of A B to B C (55.): then, because AD: CD::AB: BC,

the radius BD describe a circle cutting AB in E: A B shall be divided in extreme and mean ratio in the point E. Because (I. 36.) the squares of CA, A B are together equal to the square of CB, that is (I. 32.), to the squares of CD, D B, together with twice the rectangle CD, DB, and that the square of CA is equal to the square of CD (I. 25. Cor.); the remaining square of AB is equal to the square of D B, together with twice the rectangle CD, DB, or, which is the same thing, to the square of B E, together with the rectangle A B, BE, for BE is equal to D B, and A B to twice CD.

But (I. 30. Cor.) the square of A B is also equal to the rectangle A B, A E, together with the rectangle A B, B E. Therefore the square of B E, together with the rectangle A B, B E, is equal to the rectangle AB, AE, together with the rectangle AB, BE. Therefore the rectangle AB, AE is equal to the square of BE (1. ax. 3); and (38. Cor. 1.) AB: BEBE AE, that is, A B is divided in the point E in extreme and mean ratio. (def. 21.) Therefore, &c.

Scholium.

The parts of a line thus divided belong to a certain class of incommensurable magnitudes, described in the following general theorem.

If P, Q be two magnitudes of the same kind, and such that P is contained in Q a certain number of times with a remainder which is to P as P is to Q; the magnitudes P and Q shall be incommensurable.

For, let P be contained in Q 5 times, and let R be the remainder; then, because R: P::P: Q, 5 R: 5 P:: P: Q (17. Cor. 2.); therefore, alternando, 5 R: P::5 P: Q, and dividendo, 5R-P: P: 5P~Q: Q. But PRQP: therefore, ex æquali, 5R-PR: 5P~Q (or R): P; in which proportion the third term is less than the fourth: therefore the first term is less than the second (14.), that is, R is also contained in P 5 times, with a remainder 5R-P or R,, which is to R as R to P. Therefore, again, R, is contained in R 5 times with a remainder R, which is to R, as R, to R; and so on: i.e. every following remainder is always contained in the preceding 5 times with a new remainder, and there is no remainder which is contained in the preceding a certain number of times ex

2

actly; therefore P and Q have no common measure (5. Cor. 1.). Therefore, &c.

Now it is evident that the segments of a line medially divided are magnitudes of this description, for the greater segment E B being contained in the whole line AB once with a remainder AE which is to EB as E B to A B, it follows, as in the preceding demonstration, that AE likewise is contained in E B once with a remainder, which is to A E as AE to EB; and therefore AE, EB are incommensurable.

Another instance of incommensurables of this class is afforded by the side and diagonal of a square.

D

F

Let A B be a side, and A C one of the diagonals of the square ABCD. Produce AC to E so that CE may be equal to CB; and from CA cut off C F likewise equal to CB. Then because A E is equal to the sum, and A F to the difference of A C, C B, the rectangle under AF, AE is (I. 34.) equal to the difference of the squares of A C, CB, that is (I. 36. Cor. 1.), to the square of AB. Therefore (38. Cor. 1.), AF is to A B as AB to A E. But, because A B is equal to BC, and that E F is equal to 2 B C, AB is contained in AE twice with the remainder A F. Therefore A B, A E are magnitudes of the same kind with P, Q in the above theorem, and are incommensurable. And because A E is incommensurable with A B, and that its part CE is equal to A B, the remainder, that is, the diagonal A C, is incommensurable with A B.

We shall conclude this Scholium with an easy method of approximating numerically to all ratios of this description.

To explain it, let us suppose, as in the latter instance, that there are two magnitudes, P and Q, and that P is contained in Q twice with a remainder R, which is to P as P to Q: the ratio of P to Q shall lie between any two consecutive ratios of the series 1:2, 2:5, 512, 12:29, 29: 70, &c. the terms of which are formed from the two first, by making every new term equal to twice the last, together with the last but one; that is, it shall lie between 1:2 and 2:5; again between 2: 5 and 5: 12, &c., the ratios of each successive pair approaching always more nearly to one another, and therefore to the ratio sought,

For, since P is contained in Q twice with a remainder R, Q=2 P+R: therefore R: PP:2 P+R; that is, a ratio the same with that of R to P is formed by taking P for a new antecedent, and twice P together with R for a new consequent.

The same may be shown of the ratio 2 P+R: 5P+2 R, which is formed from this last according to the same rule, viz. that it is likewise the same with the ratio from which it is formed, that is, with the original ratio; and so on. Let these successive ratios be written one after another in the order in which they are derived: (a) R: P; (b) P: 2 P + R, (c) 2 P+R: 5 P+2R; (d) 5 P+ 2R: 12 P+5 R; (e) 12 P +5 R: 29 P +12 R, &c. Now, of these ratios, each of which is equal to the ratio sought, the ratio (c) lies between (23. Cor. 3.) 2 P: 5 P and R: 2 R, that is (def. 5.) between 2:5 and 1:2; in like manner, (d) lies between 5 12 and 2: 5, (e) between 12 : 29 and 5: 12, &c. Again, 5: 12 is formed by adding the terms of 2 × 2: 2 x 5 to those of 1: 2, and therefore lying between 2: 5 (1. Cor. 3.) and 1:2 approaches more nearly to 2:5 than 1:2 approaches; in like manner, 12: 29 is formed by adding the terms of 2x52x12 to those of 2: 5, and therefore lying between 5: 12 and 2: 5, approaches more nearly to 5:12 than 2:5 does; and so on. Not only, therefore, does the ratio sought lie between the successive ratios of the series 1:2, 2:5, 5 12, 12: 29, &c. ; but these approach continually more and more nearly to one another, and therefore to the ratio sought.

In the same manner, if P were contained in Q 5 times, we might approximate to their ratio by means of the series 1, 5, 26, 135, &c. which is formed from the two first terms by making every following term equal to 5 times the last, together with the last but one: whence the application of the Rule to any given case is sufficiently apparent.

In the case of the medial ratio the series becomes 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, &c. which is formed from the two first terms by making every following term equal to the sum of the two last terms; and this is the simplest case possible.*

The converging ratios in the text are also derivable from the doctrine of Continued Fractions. The series are of the kind called recurring series, because every term has a given relation to a certain number of the terms preceding it. It may also be remarked,

PROP. 60. Prob. 10.

D

Upon a given hypotenuse A B, to describe a right angled triangle, which shall have its three sides proportionals. Divide AB medially (59.) in the point C, and from C draw CD (I. 44.) perpendicular to AB; bisect A B in E, and from the centre E, with the radius EA describe a circle cutting CD in D; join DA, D B, DE; ADB shall be the triangle required.

A CE B

For the triangle ADB is right-angled at D, because ED is equal to EA or EB (I. 19. Cor. 4.). Also A D is to D B as the rectangle (I. 36. Cor. 2.) under A C, A B, to the rectangle under CB, AB, that is, as AC to CB (35.); and D B is to B A as the rectangle under CB, BA, to the square of B A, that is (35.) as CB: BA. But because A B is medially divided in C, AC: CB::CB: BA; therefore (12.) A De: D B : : D B: BA, and AD: DB::DB: BA (37. Cor. 4.); that is, AD, DB, BA are proportionals. Therefore, &c.

PROP. 61. Prob. 11.

Through a given point A, to draw a straight line such that the parts of it intercepted by the legs, B C, BD, of a given angle B, may be to one another in

a given ratio.

Let the given ratio be that of the two straight lines P, Q, of which the latter is greater than the distance of A from BE. From A to BD draw the straight line AE equal to Q, and take AF equal

to BE, take any line Q' which is not less than that perpendicular, and P such that P': Q':: P: Q (53.), and proceed with P', Q', as above.

PROP. 62. Prob. 12.

Through a given point P, to draw a straight line which shall pass through the intersection of two given straight lines AB, CD, that intersection being without the limits of the draught.*

Through P draw any straight line cutting the given lines in the points A, C; and through

y point B in AB draw BD parallel to AC to meet CD in D. (I. 48.) Take Pa equal to twice PA, and Pc equal to twice P C: join a B, e D, and let them be produced to meet one another in Q; and join PQ; PQ shall be the line required.

D

D

B

E

For, let BD cut P Q in E; then, because BD is parallel to a c, B E is to ED as a P to Pc (30.), that is, as AP to PC (17.); and therefore the line PQ produced (30.) passes through the intersection of A B, C D. Therefore, &c.

to BD, the rectangle under BF, BE is to the rectangle under BA, BD as the square of B E to the square of B D (37. Cor. 1. and 2.), that is, as BC to BD. But the rectangle under A B, B C is likewise to the rectangle under A B, B D, as BC to BD (35.); therefore the rectangle under EB, BF is equal to the rectangle under AB, BC (11. Cor. 1.). Therefore, &c.

PROP. 64. Prob. 14.

Given any number of straight lines A, A,, A ̧, A., antecedents, and as many B, B., B., B4, consequents, to find a straight line such that A shall bear to it the ratio which is compounded_of the ratios of A to B, A to B2, A, to B3, and A, to B..

Find (53.) the straight line P such that B: PA, B2, Q such that P: Q :: A, B, and R such that Q: R:: A,: B.; R shall be the straight line required.

For the ratio of A to R is compounded (def. 12.) of the ratios of A to B, B to P, P to Q, and Q to R, that is, of the ratios of A to B, A, to B2, A, to B., and A to B4.

Therefore, &c.

PROP. 65. Prob. 15. (Euc. vi. 18.)

describe a rectilineal figure similar to a Upon a given straight line A B, to given rectilineal figure CDEFG.

Method 1. Join CE, CF; at the points A, B (I. 47.) make the angles BAK, ABK, equal to the angles

M

DCE, CDE respectively; and because these two angles are, together, less than two right angles (I. 8.) AK, BK will meet, if produced, in some point K (I. 15. Cor. 4.). Again, at the points A, K, make the angles KAL, AKL equal to the angles ECF, CEF respectively, and (as before) let A L, KL meet one another in the point L: lastly, at the points A, L make the angles LAM, ALM equal to FCG, CFG respectively, and let A M, LM meet one another in the point M: the rectilineal figure AB KLM shall be similar to CDEFG.

For the angles of the two figures are evidently equal, each to each, because they are the sums of corresponding