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triangles FH B, G K F, which have their But if C is to be opposite to one of angles equal to one another, each to each, the given sides as A, take CE equal to because their sides are parallel (18.): B as before: from the centre E, with a therefore (5.), their other sides are also radius equal to A, describe a circle equal to one another, and F B is equal cutting Ĉ D in the points D, D; and to G F. In like manner it may be shown, join ED, ED. Then, if the points D, D that each of the other five parts into fall both on the same side of C with the which AB is divided, is equalto FB or GF. angle, each of the triangles EC D will
And it is evident that, by a similar satisfy the given conditions; but if they process, A B may be divided into any fall upon different sides, only one of other number of equal parts.
them, as ECD, will satisfy those conTherefore, &c.
ditions, and therefore that one will be It is convenient to draw the parallels the triangle required. DF, E G, &c. by making the angle A B a, · If the points D, D, being upon the equal to B A C (47.), taking B d, de, &c. same side with the angle, coincide with each equal to M, and joining correspond- one another, which will happen when ing points D d, Ee, &c. For the EDC is a right angle (12. Cor. 3.), there straight lines which join the extremities is only one triangle EDC, which is the of equal and parallel straight lines, are triangle required. also themselves equal and parallel (21.). Case 3. Let the given parts be three A similar process may be recommended, sides, viz. A, whenever a number of parallels are to B, and C D. Abe drawn without the assistance of a From the cen- B parallel ruler.
tres C, D,with
radii equal to PROP. 50. Prob. 9. (Euc. i. 22).
A, B, respectively, describe circles interToconstruct a triangle from three parts secting in E, and join EC, ED: then given, of which one, at least, is a side. ECD is evidently the triangle required. Case 1. Let
Therefore, &c. the given parts be two angles
Scholium. a side ; viz. A
In each of hese three cases we the angles A
have supposed the data, or given parts, B, and the side
to be such that the problem is possible : CD.
and the same will be supposed in all future If the given
problems. Many of them, however, will side is to be
be found to be of that description, that, interjacent
if the relations of the data be not conto lie between the given angles, make fined within certain limits, the required (47.) at the points C, D, the angles DCE, solutions will be impossible. In proCDE, equal to A, B, respectively : and blems of a more complex nature than the triangle E C D will evidently be the the present, the determination of these triangle required.
limits is facilitated by the consideration But if CD is to be opposite to one of of loci, to be noticed hereafter in the given angles, as B, make at the point Book III. Sect. 6), and which may here C, the angles DCE, ECF, equal to be explained as circumscribing the A, B, respectively, and through D, draw range, if it be limited, within which DE parallel to C F(48.): then the angle every particular datum confines the soluCE D is (15. Cor. 2.) equal to the alter- tion of the problem ; for it is obvious pate angle ECF, that is, to B, and that, if a second datum' cannot be satisECD is the triangle required.
fied within that range, the two will be Case 2. Let the given parts be two inconsistent, and the solution impossible. sides and an angle ; viz. the sides A, B, In the mean time, it may be oband the angle C.
served, the limits of the data are in If the given angle
many cases readily suggested by the is to be included,
known properties of the figure, or by take CD, CE, upon
the necessary construction of the proits sides, equal to
blem. It is evident, for instance, that in A, B, respectively,
every case of the problem before us, there and join D E; the
are certain limits, beyond which the triangle C D E will
solution will be impossible. This will evidently be the tri
happen in the first case if the two given angle required.
angles should be together equal to or
greater than two right angles, for there has its opposite sides equal to one anocan be no triangle having two of its an- ther, it is a parallelogram (22.); and it gles together equal to or greater than has two sides and the included angle the two right angles (8.). The problem will same with those given; therefore, it is be impossible in the second case, if the parallelogram required. when the side A is to be opposite to the Therefore, &c. given angle C, it should be less than the N. B. The same end will be obtained perpendicular let fall from E upon CD by drawing parallels to CB, CA (12. Cor. 4.); or again, if the given angle through the points A, B. being right, or obtuse, the side A, which It may be observed that if AC be is to be opposite to it be less than the equal to BC, the parallelogram is a side B, for no triangle can have the less rhombus; if AC be at right angles to side opposite to the greater angle (9.). BC, it is a rectangle; and if AC be And in the third case, the problem will both equal to BC, and at right angles be impossible, if every two of the given to it, it is a square. This problem, sides be not greater than the third side therefore, includes the following as parti(10.).
cular cases; 1st, to describe a rhombus
with a given side and angle; 2d, to dePROP. 51. Prob. 10.
scribe a rectangle with two given sides ; Given the three angles of a triangle and 3d, to describe a square upon a (together equal to two right angles), given finite straight line. and the perimeter AB, to construct the triangle.
PROP. 53. Prob. 12. At the points A,
To describe a triangle which shall be B, make (47.) the
equal to a given quadrilateral ABCD, angles BAC, ABC,
and shall have a side and angle adjacent equal to two of the
to it, the same with a given side AB given angles, each
and adjacent angle B of the quadrito each; bisect (46.)
lateral. the angles at A, B,
Join AC: through by the straight lines A c, B c, meeting D (48.) draw D E in c, and through c draw (48.) ca, cb, parallel to A C, to parallel to CA, CB, respectively: the meet B C produced triangle ABC shall have its angles equal in E, and join A E. to the three given angles, and its peri- Then, because DE meter equal to the given perimeter A B. is parallel to AC, The first is evident, for its sides are pa- the triangle ACE is equal to ACD rallel to the sides of the triangle CAB, (27.): therefore A B C, A CE together which has its angles at A and B, and
are equal (ax. 2.) to ABC, ACD totherefore at C, equal to the given angles gether: that is, the triangle ABE is (18.). Again, because ca is parallel to equal to the quadrilateral A B C D; and CA, the angle acA is equal (15.) to the it has the same side A B, and adjacent alternate angle C Ac, that is to the angle angle B with the quadrilateral. a A c, and therefore (6.) ac is equal to aA; А.
Therefore, &c. and in the same manner it may be shown that bc is equal to 1 B; there
PROP. 54. Prob. 13. fore, the three sides of the triangle abc are together equal to A B, that is, to the be equal to a given rectilineal figure
To describe a triangle which shall given perimeter.
ABCDEF, and shall have a side and Therefore, &c.
adjacent angle the same with a given
side AB and adjacent angle B of the PROP. 52. Prob. 11.
figure. Given two sides and the included
Join A C, AD, A E: through F draw angle of a parallelogram, to construct FG parallel to A E (48.), to meet D E the parallelogram.
produced in G: through G draw G H Let A C, B C, be the
parallel to AD, to meet C D produced in two given sides, and C
H; through H draw HK parallel to the given angle. From
HC, to meet B C produced in K, and the centres A, B, with
join A K, AH, A G.* radii equal to B C, AC,
Then because the triangle AD G is respectively, describe circles intersecting in D, and join D A, D B.
* A H is not joined in the figure, to avoid consu
sion : for the same reason, AM and AL are not Then because the quadrilateral AD joined in the correspouding figure of Prop. 55.
equal to the quadrilateral A DE F(53.), given figure, and the latter is bisected and that the triangle ADH is equal to by the straight line AN. AD G (27.), the triangle A DH is equal Therefore, &c. to the quadrilateral ADE F. To each Cor. 1. By a similar construction, of these equals add the quadrilateral any part required, for example a fifth, ABCD; therefore the quadrilateral may be cut off from a given triangle ABCH'is equal to the whole figure or rectilineal figure, by a straight line ABCDEF. But again (53.) the tri- drawn from one of its angles. For angle A B K is equal to the quadrilate- in the case of the triangle, if the base ral ABCH: therefore the triangle B C be divided into five equal parts
in the points D, &c. the triangles A BD, &c. will be equal to one another (27.) ; and therefore any one of them, as ABD, will be equal to one fifth of the given triangle. And hence the passage is easy to the division of the rectilineal figure; for it is evident (53.) that, what
ever be the point L taken in A K, the A B K, is equal to the figure ABCD figure A B C D N, constructed as above, EF, and it has the same side A B, and will be equal to the triangle A B L, and adjacent angle B with the figure. will therefore be a fifth part of the given Therefore, &c.
figure, if A B L be a fifth of the triangle
It is manifest in this case, that, if the To bisect a given triangle or rectili- given fractional part be such, that B L neal figure by a straight line drawn is less than BC, the problem will be from a given angle.
solved at once by joining A L. First, let it be required to bisect the Cor. 2. And hence it appears in what triangle A B C by a straight line to be manner a triangle or rectilineal figure drawn from the angle A.
may be divided into any number of equal Bisect (43.) B C in D,
parts by straight lines drawn from one of and join AD. Then be
its angles. cause B D is equal to DC,
PROP. 56. Prob. 15. the triangle A B D is (27.) equal to ADC, and the
To bisect a given triangle or rectilitriangle A B C is bisected
neal figure by a straight line drawn by the straight line AD.
from a given point in one of its sides. Next, let it be required to bisect the
First, let it be required rectilineal figure ABCDEF by a
to bisect the triangle straight line, to be drawn from the ABC, by a straight line angle A.
to be drawn from the With the same side A B and angle B, point D in the side A B. describe (54.) the triangle A BK, equal
Join DC: bisect (43.) to the given rectilineal figure : bisect A B in E:
through E draw (48.) E F, (43.) B K in L: through L draw (48.) parallel to DC, and join DF, EC. Then, L M parallel to A C, to meet C D pro
because D C is parallel to E F, the triduced in M: through M draw M N angle D E F is (27.) equal to CEF; parallel to AD, to meet D E in N, and therefore D EF, BEF together, are
equal to CEF, B E F together, that is, the triangle D B F, is equal to the triangle C EB, or (27.) to half of the tri. angle ABC. Therefore, ABC is bisected by the straight line D F.
Next, let it be required to bisect the
figure A B C D E F, by a straight line, to join AN, AM, AL. Then, it may be be drawn from the point G in the side shown, as in the last proposition, that A B. the triangle A B L is equal to the figure ABCDN; but A B L is equal to half the triangle ABK, that is to half of the given figure; therefore the figure ABCDN, is also equal to half of the
Describe the triangle A BK (54.) Cor. 1. Hence, upon a given base, a equal to the given figure, and, as al- rectangle may be described which shall ready shewn, bisect it by the straight be equal to a given rectilineal figure: for line GL, drawn from the point G: a triangle (54.) may be described equal join GC, GD, &c.; through L draw to the figure, and a rectangle equal to (48.) LM, parallel to GC, to meet the triangle. CD produced in M: through M draw Cor. 2. It is evident that, with a like M N parallel to G D to meet DE in construction, a parallelogram may be N, and join GN, GM, G L.* Then, described upon a given base B D which as in the last proposition, it may be shall be equal to a given triangle A B C, shown that the figure GBCDN is , and shall have one of its angles equal to equal to the triangle G B L, that is to a given angle B D E. half of the given figure; therefore the
PROP. 58. Prob. 17. (Euc. ii. 14). latter is bisected by the straight line GN.
To describe a square which shall be Therefore, &c.
equal to a given rectangle A B C D. Cor. 1. By a similar construction,
Produce A B to E,
so that B E may be any part required may be cut off from a given triangle or rectilineal figure, by a equal to B.C: bisect straight line drawn from a given point
A Ein F (43.) ; from in one of its sides. For we have but to the centre F, with make CB E the same part of C B A in the radius FA or the first case, and G B L the same part FE, describe a circle, and produce C B of A B K in the case of the rectilineal to meet the circumference in G : the figure, and proceed as before.
square of BG shall be equal to the If the fractional part be such that rectangle A B C D. BL is less than B C, the problem will
Join FG, then, because FA is equal be solved at once by joining G L.
to FE, AB is equal to the sum, and Cor. 2. And hence a given triangle or
BE to the difference, of F E, FB: rectilineal figure may be divided into any therefore (34.) the rectangle AB, BE, number of equal parts, by straight lines that is, the rectangle A B C D, is equal drawn from a given point in one of its to the difference of the squares of È E, sides.
F B, that is to the difference of the
squares of FG, F B, or (36. Cor. 1.) PROP. 57. Prob. 16. (Euc. i. 44.) to the square of B G.
Upon a given base BD to describe a Therefore, &c. rectangle which shall be equal to a given Cor. Hence a square may be described triangle ABC.
which shall be equal to a given recFrom the point D
tilineal figure (57. Cor. 1.). draw D Е, at right an.
PROP. 59. Prob. 18. gles to B D (44.):
To describe a square which shall be through A draw A E
equal to the difference of two given parallel to BD, to meet D E in E (48.): join
squares, viz. the squares of A B, A C. EB: bisect B C in F (43.): through F
From the point C draw draw FG, parallel to D E, to meet B E (44.) C D perpendicular to
A B, and from the centre in G: through G draw H K parallel to BD, and complete the rectangle DH: Scribe the circle B D cut
A with the radius A B, deDH shall be the rectangle required.
Complete the rectangles DL, DM. țing CD in D: the square of CD shall Then, because GM, GD are comple- of AB, A C.
be equal to the difference of the squares
For (36. Cor. 1.) the ments of the rectangles FH, K L, which are about the diagonal of the rectangle square of CD is the difference of the DM, G D is (23.) equal to GM. There- squares of A D, A C, of which A D is fore, adding F H to each, the whole equal to A B.
Therefore, &c. HD is equal to the whole M F, that is,
Cor. Hence a square may be de(26.) to twice the triangle ABF, or to the triangle A B C; and HD'is described, which shall be equal to the dif
ference of two given rectilineal figures scribed upon the given base B D.
(58. Cor.). Therefore, &c.
PROP. 60. Prob. 19. * GD and G L do not necessarily coincide, as in the figure,
To describe a square which shall be
equal to the sum of two, three, or any magnitude is greater or less than a number of given squares, viz. the fourth. squares of A, B, C, &c.
The ratio of one magnitude to Take D E equal to A ; from the point another is independent of the kind of D, draw DF (44.) at right angles to magnitudes compared; for it is obvious DE, and equal to B: join EF: from that one may contain the other, or the F, draw F G at right angles to E F, and sixth, or twelfth, or hundredth part of equal to C; and so proceed for the rest the other the same number of times,
; of the given squares. Join E G. Then, whether they be lines, or surfaces, or because E D F is a right angled triangle, solids, or again, weights, or parts of the square_ of EF is equal to the duration. squares of E D, D F (36.), that is, to the It is required only for the comparison squares of A, B: again, because EFG we speak of, that the magnitudes be of is a right angled triangle, the square of the same kind, containing the same E G is equal to the squares of EF, FG, magnitude, each of them, a certain numthat is, to the squares of A, B, C; and ber of times, or a certain number of
times nearly. Upon these numbers, and upon these only, the ratio depends.
Hence it appears that this theory pertains in truth to Arithmetic. The use of Proportion is, however, so indispensable in Geometry, that it has been usual, either to introduce its theorems into the body of the science, after the
example of Euclid, or to premise a few Therefore, &c.
of the most important of them as a Cor. Hence a square may be de- manual for reference. In the present, scribed, which shall be equal to the sum and two following sections, the subject of two, three, or any number of given will be discussed at a length commenrectilineal figures (58 Cor.).
surate with its importance.
Def. 1. When one magnitude is comBOOK II.
pared with another of the same kind, § 1. Ratios of Commensurable Magni- the first is called the antecedent, and the
tudes-9 2. Proportion of Commen. second the consequent. surable Magnitudes --- 3. General 2. One magnitude is said to be a Theory of Proportion-$4. Propor. multiple of another, when it contains tion of the sides of Triangles—5. that other a certain number of times Proportion of the surfaces of Recti. exactly : and the other magnitude, which lineal Figures — 6. Properties of is contained in the first a certain numLines divided Harmonically —87. ber of times exactly, is said to be a subProblems.
multiple, or measure, or part of the first.
Hence, also, one magnitude is said to SECTION 1. Ratios of Commensurable
measure another when it is contained in Magnitudes.
the other a certain number of times exIn the language of Mathematics, the actly. Latin word ratio has been adopted to 3. Two magnitudes are said to be express what is more generally under- equimultiples of two others, when they stood by the term proportion : thus, contain those others the same number instead of “the proportion which” one of times exactly: and the other magnithing bears to another, we say “the tudes which are contained in the first ratio which ”. one bears to the other, the same number of times exactly, are meaning its comparative magnitude- said to be like parts of the two first. instead of saying that A is to B “in Thus, 7 A, 7 B, are equimultiples of the proportion of 5 to 6," we say "in A, B ; and A, B, are like parts of 7 A, the ratio of 5 to 6."
7 B. The word proportion has
4. Two magnitudes are said to be other hand been appropriated to ex- commensurable with one another, when press the equality of ratios, as here- a common measure of the two may be after defined; or, as it may be here found, i. e. a magnitude which is conless minutely explained, the case in tained in each of them a certain number which one magnitude is as many times of times exactly. greater or less than another, as a third In like manner, any number of mag