angles of equiangular triangles. Also, equal to CDEFG, may then be easily because the triangles KBA, EDC are derived from I. 21. similar, B K is to KA as D E to EC; Therefore, &c. and for the like reason, KA is to KL The last method will as EC to EF: therefore, ex æquali, appear under a still BK:KL::DE: EF. And in the

more simple and convesame manner it may be demonstrated, nient form, if A B be that the sides about the other equal made to coincide with angles of the two figures are propor. CD, and B with D, as the adjoined tionals. Therefore, the figure described figure. upon A B is similar to the given figure CDEFG. (def. 14.)

PROP. 66. Prob. 16. Method 2. Place A B parallel to CD, To describe a rectilineal figure, which and join CA, D B. Then, if A B be shall be similar to a given rectilineal not equal to CD, CA will not be pa- figure CDEFG, and shall have its perirallel to D B (I. 21. and 14. Cor. 2.), meter equal to a given straight line AQ. and CA, D B will meet, if produced, in Produce CD to R, (see the second some point P: join P E, PF, PG: figure of Prop. 65.) so that CR may be through B draw B K parallel to DE equal to the perimeter of the given (I. 43.), and let it meet P E in K: figure CDEFG; take A B (53.) a through K draw K L parallel to EF, fourth proportional to CR, A Q, and and let it meet PF in L: through L CD, and upon AB (65.) describe a draw L M parallel to F G, and let it meet rectilineal figure A B K L M similar to PG in M; and join AM: the figure the given figure. Then, because the A BK L M shall be similar to the given figures AB KLM and CDEFG are figure CDEFG.

similar, the perimeter of the first is to the
perimeter of the other (43.), as AB to
CD, that is (12.), as AQ to CR: but
the perimeter of C D E F G is equal to
CR: therefore (18. Cor.) the perimeter
of A B K L M is equal to A Q, that
is, to the given perimeter. Therefore,
AB K L M is the figure required.

Therefore, &c.
PROP. 67. Prob. 17. (Euc. vi. 25.)

To describe a rectilineal figure which shall be similar to a given rectilineal figure ABC, and shall have its area

equal to the area of another given rectiFor, because PC:PA:: PD:PB lineal figure DEF. (29.), that is, :: PE: PK, that is :: Find (1.58. Cor.) PF: PL, that is, :: PG:PM, A Mis the straight lines parallel to CG; and because the sides G, H such that of the two figures are parallel, each to their squares may each, they are equiangular (I. 18.) with be equal to the tione another.

Also, because the tri- gures ABC, DEF angles CDP, A B P are similar, CD: respectively : take DP:: A B :B P, and for the like reason KL a fourth proDP:DE:: BP: BK; therefore, ex portional (53.) to æquali, CD:DE:: AB: B K. And G, H, and BC, in the same manner it may be demon- and upon KL (65.) describe the figure strated, that the sides about the other KLM similar to the figure BCA. equal angles of the two figures are pro- Then, because the figure A B C is to the portionals. Therefore the figure upon figure MKL as the square of BC A B is similar to the given figure. When to the square of K L (43.), that is, (37. A B is equal to CD, CA is parallel to Cor. 4.) as the square of G to the square BD, and the lines EK, FL, GM, in- of H, and that the figure A B C is equal stead of being drawn to a point P, must to the square of G, the figure K L M is be drawn parallel to C A or B D. The (18.) equal to the square of H, that is, demonstration that the figure ABKLM, to the figure DEF; and it is similar to so constructed, will be both similar and ABC; therefore it is the figure required,

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PROP. 68. Prob. 18.

as before, the part PBCDL of the

given figure, is equal to the triangle To divide a given rectilineal figure PBH, and the remaining part PAFEL ABCDEF in a given ratio, by a to the triangle PH G'. Therefore, the straight line drawn from one of its an parts of the figure are as the bases B H, gles, or from a given point in one of HG' (39.), that is, in the given ratio. its sides.

Therefore, &c. In the first place, let A be the given angle from which the line of

sion is to

PROP. 69. Prob. 19. be drawn.

Given a triangle A, a point B, and two straight lines CD, CE, forming an angle D CE; to describe a triangle which shall be equal to the triangle A, 80 that it may have the angle D C E for one of its angles, and the opposite side

passing through the point B. Describe (I. 54.) the triangle A BG, have three cases to consider; first, when

In the solution of this problem we equal to the figure ABCDEF, and naving the side A B, and angle A' B C the point B is in one of the given lines the same with it: divide (55.) the base without the given angle D'CE; and

as CD; secondly, when the point B is BG in the given ratio in the point H: join AC and AD: through H draw thirdly, when B is within the angle DCE. H K parallel to A C, to meet CD pro- CD. Take CD equal to a side of the tri

Case 1. Let the point B be in the line duced in K; through K draw KL rallel to AD to meet D E in L, and angle A, and upon CD (1.50.) describe join AL: AL shall be the line re

a triangle CDF, which shall have its two quired.

remaining sides equal to the two remain. For, by the construction, which is si- ing sides of the triangle A, each to each, milar to that of 1.55., it is evident, that and therefore (I. 7.) shall be equal to the the figure ABCDL is equal to the triangle A BH:* but the whole figure is equal to the triangle ABG: therefore, the part A LEF is equal to the triangle AHG. Therefore, the parts of the figure are as the triangles A BH, AHG, that is (39.) as the bases BH, HG, that is, in the given ratio.

triangle A in every respect: through F Next, let P be the given point in the (I. 48.) draw F E paraīlel to DC: join side A B, from which the line of division DE, BE: through D (I. 48.) draw DG is to be drawn.

parallel to BE to meet CE in G, and join GB: the triangle CBG shall be the triangle required.

For, because E B is parallel to GD, the triangle GEB (I. 27.) is equal to DEB: therefore, adding the triangle ECB to each, the whole triangle GCB is equal to ECD, that is (because E F is

parallel to CD) to FC D, that is, to the Describe, as before, (I. 54.), the tri- given triangle A. angle ABG equal to the given figure, Case 2. Let the and, by drawing A G' parallel to PG; point B be withmake the triangle PB G’equal to ABG out


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the angle (as in I. 56.): divide (55.) B G' in the DCE. Through given ratio in the point H: join PC B (I. 48.) draw and PD: through Å draw H K paral- B D parallel to lel to PC, to meet CD produced in K: CE to meet CD through K draw K L parallel to PD to in D, and by the construction pointed out meet DE in L, and join P L. Then, in Case 1. describe a triangle DCF

having the given side D C and the given • The line AH is wanting in this figure; and PH is in like manger wanting in the figure below, angle DCE, and equal to the given trias also AG, A G'.

angle A: in CF produced take the




point G such that CGXGF may be which shall form equal to B DXC F (56.): join BG, and with two given let BG cut C D in H: CHG shall be straight lines CD the triangle required.

CE cutting one Join HF. Then, because CG XGF another in a is equal to B DX CF, CF: FG::CG triangle equal to : BD (38.), that is, since DHB and a given triangle


C HD h 11 CHG are (I. 15.) equiangular, :: CH A ;" we should : HD. Therefore (29.) H F is parallel have had four to DG; and hence, as before, the trian- solutions, two gles DHF and GHF are (I. 27) equal to of them correone another, and CHG is equal to CDF, sponding to the that is to A.

angle in which Case 3. Let the

DB lies, and une point B be within

for each of the adjacent angles; as is the angle DCE.

apparent from the foregoing construcThrough B (I. 48.)

tions applied to the adjoined figure. draw BD parallel to

When B D is equal to a fourth of CF, EC to meet CD in

two of these become identical; when D, and, by the con

BD exceeds a fourth of CF, the same struction pointed out

two become impossible, but the other in Case i, describe

two, viz. those which correspond to the a triangle D CF, having the given side adjacent angles, are always possible, exCD and the given angle DCE, and equal cept when B is in one of the lines as to the given triangle A : in CF (if it be cd. This last-mentioned position is possible) take the point G such that peculiar : it has likewise, however, two CGXGF may be equal to BDXCF solutions, one for each of the adjoining (56. Cor.): join BG, and let GB pro- angles. duced cut C D in H; CHG shall be the triangle required.

BOOK III. Join HF. Then, by a demonstration which may be given in the same words ♡ 1. First Properties of the Circleas that of Case 2, the triangle CHG is

$ 2. Of Angles in a Circle-3. Rectequal to CDF, that is to the given tri

angles under the segments of Chords angle A.

- 4. Regular Polygons, and ApIn this last case a solution will be

proximation to Circular Area—55.

Circle a Maximum of Area, and impossible if BD exceed a fourth of

Minimum of Perimeter-$ 6. Simple CF; for then BDXC F will exceed a fourth of the square of CF, that is

and Plane Loci— 7. Problems. (I. 29. Cor. 2.) the square of half CF;

Section 1.–First Properties of the and no point G can be taken in CÉ

Circle. such that CGXGF may exceed the square of half C F (56. N.'B.).

Def. 1. Any portion of the circumWe may remark also that, whenever ference of a circle is called an arc; and the solution is possible in the last case, the straight line which two points G may be found such that CG joins the extremities of x GF = BD XCF, (56.) and therefore an arc is called the chord two lines G H may be drawn satisfying of that arc. the given conditions. When B D is ex

When the chord passes actly a fourth of C F, these two points through the centre, it is coincide with one another and with the a diameter, and the arcs middle point of C F, and therefore the upon either side of it, being equal to two solutions become identical.

one another, are called, each of them, If В be in one of the lines DC, EC a semi-circumference.* produced, or within the angle which is 2. The figure which is contained by vertical to DCF, the solution will be an arc and its chord is called a segment. manifestly impossible.


diameter divides a circle and its cir Scholium.

cumference into two equal parts, is evident from the

symmetrical character of the circle. The same may Had the problem been proposed be proved by doubling the figure upon the diameter under the following form : “through for every point of the cirenmference being at the a given point B to draw a straight line (whether of circumference or area) will coincide

• That

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A segment which is contained by a semi

Prop. 1. (Euc. iii. 2.) circumference and a diameter, is equal to half the circle, and is therefore called

If a straight line meet the circuma semicircle.

ference of a circle in two points, it shall 3. An angle in a segment is the

out the circle in those points ; und the angle contained by two

part of the straight line which is straight lines drawn

between them shall fall within the

circle. from any point of the arc of the segment to

Let the straight line AB meet the the extremities of its

circumference of a circle having the base or chord.

4. A sector of a circle is the figure con. tained by any arc, and the radii drawn to its extremities.

5. Equal circles are those which have equal radii.

centre C in the points A and B: it By this is intended no more than that shall cut the circle in those points, and when the term “ equal circles" may be

the art AB shall fall within the

circle. hereafter used, those which have equal radii are to be understood. That such

Join CA, CB : bisect A B in D, and circles, however, have also equal areas, in the line A B, the former between A

join C D : take also the points E and F is at once evident by applying one to and B, the latter upon the other side of the other, so that their centres may B, and join CE, CF. coincide. 6. Concentric circles are those which CAB is an isosceles triangle ; there

Then, because CA is equal to CB, have the same centre.

fore (1. 6. Cor. 3.) C D, which is drawn 7. (Euc. iii

. def. 2.) A straight line is from the vertex to the bisection of the said to touch a circle, when it meets the base, is perpendicular to A B. And, circumference in any point, but being because CE is nearer to the perpendiproduced does not cut it in that point. cular than C B is, (1. 12. Cor. 2.) CE Such a line is frequently, for brevity's is less than CB; therefore the point Ę sake, called a tangent; and the point is within the circle: on the other hand, in which it meets the circumference is because C F is farther from the perpencalled the point of contact.

dicular than C B is, C F is greater than

CB; and therefore the point F is with8. Circles are

out the circle. And the same may be said to touch one

said of the parts about A. Therefore another, when they

the straight line in question cuts the cirmeet, but do not

cle in the points A and B. cut one another.

Also, because every point, as E, of

C than 9. A rectilineal figure is said to be A B, is at a less distance fror inscribed in a circle, when all its angular the radius, the part A B falls within points are in the cir

the circle, cumference of the cir

Therefore, &c. cle. Also, when this

Cor. 1. A straight line cannot meet a is the case, the circle

circle in more than two points, is said to be circum

Cor. 2. A straight line which touches scribed about the rec

a circle meets it in one point only. tilineal figure.

Cor. 3. A circle is concave towards 10. A rectilineal figure is said to be its centre. circumscribed about a circle, when all

PROP. 2. (Euc. iii. 16.) its sides touch the circle. Also, when

The straight line, which is drawn at this is the case, the

right angles to the radius of a circle circle is said to be

from its extremity, touches the circle ; inscribed in the rec

and no other straight line can touch it tilineal figure.

in the same point,

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conversely, if a diameter bisect any A in the circum

other chord, it shall cut it at right ference of a circle

angles. having the centre

Let AB be a diameter of the circle C, let the straight

ADE, the centre of line A T be drawn

which is C; and, at right angles to

first, let it cut the the radius CA:

chord D E at right AT shall touch

angles in the point the circle.

F: DE shall be For, C A being shorter (I. 12. Cor. 3.) bisected in F. than any other line which can be drawn Join CD, CE. Then, because CDE from C to A T, every other point on is an isosceles triangle, the straight line AT lies without the circle: therefore C F which is drawn from the vertex C AT meets the circle in the point A but at right angles to the base D Е, bisects does not cut it, that is, it touches the the base, that is, DF is equal to circle.

FE. (I. 6. Cor. 3.) In the next place, let D E be any Next let D E be bisected in F by the other straight line passing through the diameter A B; the angles D FA, E FA same point A: DE shall cut the circle. shall be right angles. For in the

For, C A not being perpendicular to isosceles triangle CDE, the straight DE, let CE be perpendicular to it. Then, line which is drawn from the vertex C because CE is less than CA (I. 12. to the middle point of the base D E Cor. 3.), the point E is within the circle. is at right angles to the base. (I. 6. But if D be a point in D E upon the Cor. 3.) other side of A, C D will be farther from Therefore, &c. the perpendicular than CA; wherefore Cor. 1. A diameter bisects all chords CD being (I. 12. Cor. 2.) greater than which are parallel to the tangent at CA, the point D will be without the either extremity of the diameter. circle. Therefore the straight line DE Cor. 2. The straight line which bicuts the circle ; and the same may be sects any chord at right angles passes demonstrated of every other straight line through the centre of the circle. which passes through A, except the Cor. 3. If two circles straight line AT only, which is at have a common chord, right angles to A C.

the straight line which Therefore, &c.

bisects it at right angles Cor. 1. (Euc. iii. 18.) If a straight shall pass through the line touches a circle, the straight line centres of both the drawn from the centre to the point of circles. contact, shall be perpendicular to the Cor.4. (Euc. iii. 4.) It appears from line touching the circle.

the proposition, that two chords of a Cor. 2. (Euc. iii. 19.) If a straight circle cannot bisect one another except line touches a circle, and from the they both of them pass through the point of contact, a straight line be centre. drawn at right angles to the touching For, if only one of the chords pass Jine, the centre of the circle shall be in through the centre, it cannot be bisected that line.

by the other which does not pass Cor. 3. Tangents TA, T B which through the centre. are drawn to a circle from the same And if, when neither of them passes point T, are equal to one another. For, through the centre, it were possible that CAT, CBT being right-angled tri- they should bisect one another, the angles which have the common hypo- diameter passing through the supposed tenuse CT, and their sides CA, CB point of mutual bisection would be at equal to one another, their remaining right angles to each of them, which is sides TA, T B are likewise equal. (I. 13.) absurd.

Cor. 4. Tangents which are at the extremities of the same diameter are

PROP. 4. (Euc. iii. 15.) parallel to one another.

The diameter is the greatest straight

line in a circle; and, of others, that Prop. 3. (Euc. iii. 3.)

which is nearer to the centre is greater If a diameter cut any other chord at than the more remote : also the greater right angles, it shall bisect it; and is nearer to the centre than the less.

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