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a system of Geometry by the following: "Four quantities are proportionals, when the first is the same multiple of the second, or the same part of it, that the third is the fourth?".

18. Point out the defect of the following definition: “Four magnitudes are proportional when equimultiples may be taken of the first and the third, and also of the second and fourth, such that the multiples of the first and second are equal, and also those of the third and fourth."

19. Apply Euclid's definition of proportion, to shew that if four quantities be proportional, and if the first and the third be divided into the same arbitrary number of equal parts, then the second and fourth willeither be equimultiples of those parts, or will lie between the same two successive multiples of them.

20. The Geometrical definition of proportion is a consequence of the Algebraical definition; and conversely.

21. What Geometrical test has Euclid given to ascertain that four quantities are not proportionals ? What is the Algebraical test?

22. Shew in the manner of Euclid, that the ratio of 15 to 17 is greater than that of 11 to 13.

23. How far may the fifth definition of the fifth Book be regarded as an axiom? Is it convertible ?

24. Def. 9, Book v. “Proportion consists of three terms at least." How is this to be understood ?

25. Define duplicate ratio. How does it appear from Euclid that the duplicate ratio of two magnitudes is the same as that of their squares ?

26. When is a ratio compounded of any number of ratios? What is the ratio which is compounded of the ratios of 2 to 5, 3 to 4, and 5 to 6?

27. By what process is a ratio found equal to the composition of two or more given ratios? Give an example, where straight lines are the magnitudes which express the given ratios.

28. What limitation is there to the alternation of a Geometrical proportion

29. Explain the construction and sense of the phrases, ex æquali, and ex æquali in proportione perturbata, used in proportions.

30. Exemplify the meaning of the word homologous as it is used in the Fifth Book of the Elements.

31. Why, in Euclid v. 11, is it necessary to prove that ratios which are the same with the same ratio, are the same with one another?

32. Apply the Geometrical criterion to ascertain, whether the four lines of 3, 5, 6, 10 units are proportionals,

33. Prove by taking equimultiples according to Euclid's definition, that the magnitudes 4, 5, 7, 9, are not proportionals.

34. Give the Algebraical proofs of Props. 17 and 18, of the Fifth Book.

3 5. What is necessary to constitute an exact definition. In the demon stration of Euc. v. 18, is it legitimate to assume the converse of the fifth definition of that Book? Does a mathematical definition admit of proof on the principles of the science to which it relates ?

36. Explain why the properties proved in Book y, by means of straight lines, are true of any concrete magnitudes.

37. Enunciate Euc. v. 8, and illustrate it by numerical examples. 38. Prove Algebraically Euc. v. 25.

39, Shew that when four magnitudes are proportionals, they cannot, when equally increased or equally diminished by any other magnitude, continue to be proportionals.

40. What grounds are there for the opinion that Euclid intended to exclude the idea of numerical measures of ratios in his Fifth Book.

41. What is the obiect of the Fifth Book of Euclid's Elements :

DEFINITIONS.

I.

SIMILAR rectilineal figures are those which have their several angles equal, each to each, and the sides about the equal angles proportionals.

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II. “Reciprocal figures, viz. triangles and parallelograms, are such as have their sides about two of their angles proportionals in such a manner, that a side of the first figure is to a side of the other, as the remaining side of the other is to the remaining side of the first."

III. A straight line is said to be cut in extreme and mean ratio, when the whole is to the greater segment, as the greater segment is to the less.

IV.

The altitude of any figure is the straight line drawn from its vertex perpendicular to the base.

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PROPOSITION I. THEOREM. Triangles and parallelograms of the same altitude are one to the other as their bases.

Let the triangles ABC, ACD, and the parallelograms EC, CF, have the same altitude, viz. the perpendicular drawn from the point A to BD or BD pro

duced. As the base BC is to the base CD, so shall the triangle ABC be to the triangle ACD,

and the parallelogram EC to the parallelogram CF.

E A F

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HGB C D Produce BD both ways to the points H, L, and take any number of straight lines , GH, each equal to the

base BC; (1. 3.) and DK, KL, any number of them, each equal to the base CD;

and join AG, AH, AK, AL.
Then, because CB, BG, GH, are all equal,
the triangles AHG, AGB, ABC, are all equal : (1. 38.)
therefore, whatever multiple the base HC is of the base BC,

the same multiple is the triangle AHC of the triangle ABC: for the same reason whatever multiple the base LC is of the base CD, the same multiple is the triangle ALC of the triangle ADC:

and if the base HC be equal to the base CL, the triangle AHC is also equal to the triangle ALC: (1. 38.)

and if the base HC be greater than the base CL, likewise the triangle AHC is greater than the triangle ALC;

and if less, less; therefore since

there are four magnitudes, viz. the two bases BC, CD, and the two triangles ABC, ACD; and of the base BC, and the triangle ABC, the first and third, any equimultiples whatever have been taken,

viz. the base HC and the triangle AHC; and of the base CD and the triangle ACĎ, the second and fourth, have been taken any equimultiples whatever,

viz. the base CL and the triangle ALC; and since it has been shewn, that, if the base HC be greater than the base CL, the triangle AHC is greater than the triangle ALC;

and if equal, equal ; and if less, less ; therefore, as the base BC is to the base CD, so is the triangle ABC

to the triangle ACD. (v. def. 5.) And because the parallelogram CE is double of the triangle ABC,

(1. 41.)

and the parallelogram CF double of the triangle ACD, and that magnitudes have the same ratio which their equimultiples

have; (v. 15.) as the triangle ABC is to the triangle ACD, so is the parallelogram

EC to the parallelogram CF; and because it has been shewn, that, as the base B C is to the base

CD, so is the triangle ABC to the triangle ACD; and as the triangle ABC is to the triangle ACD, so is the paralle

logram EC to the parallelogram CF; therefore, as the base BC is to the base CD, so is the parallelogram EC to the parallelogram CF. (v. 11.)

Wherefore, triangles, &c. Q.E.D.

COR. From this it is plain, that triangles and parallelograms that have equal altitudes, are to one another as their bases.

Let the figures be placed so as to have their bases in the same straight line; and having drawn perpendiculars from the vertices of the triangles to the bases, the straight line which joins the vertices is parallel to that in which their bases are, (I. 33.) because the perpendiculars are both equal and parallel to one another. (I. 28.) Then, if the same constri be made as in the proposition, the demonstration will be the same.

PROPOSITION II. THEOREM. If a straight line be drawn parallel to one of the sides of a triangle it shall cut the other sides, or these produced, proportionally: and conversely,' if the sides, or the sides produced, be cut proportionally, the straight line which joins the points of section shall be parallel to the remaining side of the triangle. Let DE be drawn parallel to BC, one of the sides of the triangle ABC.

Then BD shall be to DA, as CE to EA.

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Join BE, CD. Then the triangle BDE is equal to the triangle CDE, (1. 37.) because they are on the same base DE, and between the same parallels DE, BC;

but ADE is another triangle ; and equal magnitudes have the same ratio to the same magnitude;

(v.7.) therefore, as the triangle BDE is to the triangle ADE, so is the

triangle CDE to the triangle ADE: but as the triangle BDE to the triangle ADE, so is BD to DA, (VI. 1.) because, having the same altitude, viz. the perpendicular drawn

from the point E to AB, they are to one another as their bases ; and for the same reason, as the triangle CDE to the triangle ADE, so is CE to EA:

therefore, as BD to DA, so is CE to EA. (v. 11.) Next, let the sides AB, AC of the triangle ABC, or these sides produced, be cut proportionally in the points D, E, that is, so that BD may be to DA as CE to EA, and join DE.

Then DE shall be parallel to BC.

The same construction being made,

because as BD to DA, so is CE to EA; and as BD to DA, so is the triangle BDE to the triangle ADE; (VI. 1.) and as CE to EA, so is the

triangle CDE to the triangle ADE; therefore the triangle BDE is to the triangle ADE, as the triangle

CDE to the triangle ADE; (v. 11.)

that is, the triangles BDE, CDE have the same ratio to the triangle

ADE: therefore the triangle BDE is equal to the triangle CDE: (v. 9.)

and they are on the same base DE: but equal triangles on the same base and on the same side of it, are between the same parallels ; (I. 39.)

therefore DE is parallel to BC. Wherefore, if a straight line, &c. Q.E.D.

PROPOSITION III. THEOREM.

If the angle of a triangle be divided into two equal angles, by a straight line which also cuts the base; the segments of the base shall have the same ratio which the other sides of the triangle have to one another : and conversely, if the segments of the base hare the same ratio which the other sides of the triangle have to one another; the straight line drawn from the vertet to the point of section, divides the vertical angle into two equal angles.

Let ABC be a triangle, and let the angle BAC be divided into two equal angles by the straight line AD.

Then BD shall be to DC, as B A to AC.

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Through the point C draw CE parallel to DA, (1. 31.)

and let B A produced meet CE in E. Because the straight line AC meets the parallels AD, EC, the angle ACE is equal to the alternate angle CAD: (1. 29.) but CAD, by the hypothesis, is equal to the angle BAD;

wherefore BAD is equal to the angle ACE. (ax. 1.) Again, because the straight line BAE meets the parallels AD, EC the outward angle BAD is equal to the inward and opposite angle

AEC: (I. 29.) but the angle ACE has been proved equal to the angle BAD;

therefore also ACE is equal to the angle A EC, (ax. 1.) and consequently, the side A E is equal to the side AC: (1. 6.) and because AD is drawn parallel to EC, one of the sides of the triangle BCE, therefore BD is to DC, as B A to AE: (VI. 2.)

but AE is equal to AC;
therefore, as BD to DC, so is BA to AC. (v. 7.)

Next, let BD be to DC, as B A to AC, and join AD. Then the angle BAC shall be divided into two equal angles hy the straight line AD.

The same construction being made ; because, as BD 10 DC, so is B.A to AC;

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