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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 BH, gles, or from a given point in one of H G' (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 division is to

PROP. 69. Prob. 19. be drawn.

Given a triangle A, a point B, and two straight lines CD, CE, forming an angle DCE; to describe a triangle which shall be equal to the triangle A, so 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 having the side AB, 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 BCE. 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 parallel 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 remainFor, 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 ABCD L 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 parallel to DC: join side A B, from which the line of division DE, B E: 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 EB 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 Describe, as before, (I. 54.), the tri- parallel to CD) to FCD, that is, to the

given triangle A. angle A B G 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 the angle (as in I. 56.): divide (55.) B G' in the DC E. Through given ratio in the point H; join PC B (I. 48.) draw and PD: through I draw H K paral- B D parallel to é lel to PC, to meet C D 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 D E 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 manner wanting in the figure below, angle DC E, and equal to the given trias also AG, A G'.

angle A: in CF produced take the

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point G such that CGx GF may be which shall form equal to B DRC F (56.): join BG, and with two given let B G 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 C a is equal to B DRCF, CF: FG::CG triangle equal to

g : BD (38.), that is, since DHB and a given triangle T

CU'D h H CH G 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

Blies, and one point B be within

for each of the adjacent angles; as is the angle D C E.

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 CGXG F may be equal to BDXC F solutions, one for each of the adjoining (56. Cor.): join BG, and let G B pro- angles. duced cut C D in H; CH G shall be the triangle required.

BOOK III. Join HF. Then, by a demonstration § 1. First Properties of the Circlewhich may be given in the same words

$ 2. Of Angles in a Circle-93. Rectas that of Case 2, the triangle CHG is

angles under the segments of Chords equal to C DF, that is to the given tri

-> 4. Regular Polygons, and Apangle A.

proximation to Circular Area-$ 5. In this last case a solution will be

Circle a Maximum of Area, and impossible if B D exceed a fourth of

Minimum of Perimeter-$ 6. Simple CF; for then BDXC F will exceed a

and Plane Loci8 7. Problems. fourth of the square of CF, that is (I. 29, Cor. 2.) the square of half CF;

SECTION 1.--First Properties of the and no point G can be taken in CF

Circle. such that CG x G F may exceed the square of half C F (56. N. B.).

Def. 1. Any portion of the circumWe may rema 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 G F = BDXC F, (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 CF, these two points through the centre, it is coincide with one another and with the a diameter, and the arcs middle point of CF, 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 D C, 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.

* That every diameter divides a circle and its cir Scholium.

cumference into two equal parts, is evident from the

symmetrical character of the circle. Had the problem been proposed be proved by doubling the figure upon the diameter under the following form : “ through for every point of the cireumference being at the

same distance from the centre, the parts so applied a given point B to draw a straight line (whether of circumference or area) wiil coincide.

The same may

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 cut the circle in those points ; and the angle contained by two

part of the straight line which is straight lines drawn

betueen them shall fall within the

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

Let the straight line A B meet the the extremities of its

circumference of a circle having the base or chord.

4. A sector of a circle is the figure contained 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"

the part AB shall fall within the may be

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, join CD: take also the points E and F is at once evident by applying one to

in the line A B, the former between A the other, so that their centres may B, and join CE, CF.

and B, the latter upon the other side of coincide. 6. Concentric circles are those which CAB is an isosceles triangle ; there

Then, because CA is equal to C B, 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, (I. 12. Cor. 2.) CE Such a line is frequently, for brevity's is less than CB; therefore the point E 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 9. A rectilineal figure is said to be A B, is at a less distance from C than 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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From the point

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 AT 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 C DE from co to A T, every other point of is an isosceles triangle, the straight line AT lies without the circle: therefore CF 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 AB; 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 ČDE, 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 DE 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 A T 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 line, 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 C A, C B point of mutual bisection would be at equal to one another, their remaining right angles to each of them, which is sides TA, TB 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.

CN

Let A B be a dia

D shall be the centre of the circle meter of the circle

ABC. AB D, the centre of

Join A B and BC, and bisect them in which is C: let DE,

the points E and F respectively; and join FG be any two chords

DE, DF. Then, because D A B is an to which perpendi

isosceles triangle, the straight line DE, culars CH, CK are

which is drawn from the vertex D to drawn; and let the

the bisection of the base A B, is at right distance C H be less than C K: the angles to A B (1. 6. Cor. 3.); and, bediameter A B shall be greater than the cause D E bisects the chord A B at right chord D E, and the chord D E shall be angles (3. Cor. 2.), it passes through the greater than the chord F G.

centre of the circle. In the same manJoin CD, CE, CF. Then, because ner it may be shown that the straight CA is equal to CD, and C B to CE, line DF passes through the centre of the whole A B is equal to CD and C E the circle. But the only point through together: but C D and D E together which each of the straight lines D E, (I. 10.) are greater than DE: therefore D F passes, is their point of intersecA B is greater than D E. Again, be- tion D. Therefore D is the centre of cause CHD, CKF are right-angled the circle. triangles, and that C D-square is equal Therefore, &c. to C F-square, the squares of CH, À D Cor. 1. From any other point than together (I. 36.) are equal to the squares the centre there cannot be drawn to of CK, KF together: but the square the circumference of a circle more than of C H is less than the square of CK; two straight lines that are equal to one therefore the square of H D is greater another, whether the point be within or than the square of KF, that is, H D is without the circle. (Euc. iii. 7 and 8 greater than KF. And DE, FG are parts of.) double of HD, K F respectively, be- Cor. 2. It appears from the demonstracause the perpendiculars CH, C K tion that if three points A, B and C be pass through the centre (3.): therefore given which are not in the same straight D E is greater than F G.

line, a circle may be found, the circumNext, let the chord D E be greater ference of which shall pass through the than FG; it shall also be nearer to the three points A, B and C; the circle, centre. For, C H and C K being drawn namely, which has for its centre the inas before, the squares of CH, HD tersection of the two lines which bisect together are equal to the squares of A B and B C at right angles. CK, KF together; but the square of HD, which is half of D E, is greater

PROP. 6. than the square of KF, which is half If two circles have the same centre, of FG: therefore the square of CH is either they shall coincide, or one of them less than the square of Ć K, and C H is shall fall wholly within the other. less than C K, that is, D E is nearer to For if the radii of two the centre than F G is.

concentric circles be Therefore, &c.

equal to one another, it Cor. (Euc. iii. 14.) Equal straight is manifest that every lines in a circle are equally distant from point in the circumthe centre; and those which are equally ference of the one must distant from the centre are equal to one

be at the same distance another.

from their common cen

tre, with every point in the circumPROP. 5. (Euc. iii. 9.)

ference of the other; and therefore the If a point be taken, from which to the two circumferences cannot but coincide, circumference of a circle there fall more But if the radii he unequal, every point than two equal straight lines, that point in the circumference of that which has shall be the centre of the circle.

the lesser radius is at a less distance Let ABC be a circle,

from the common centre, and therefore and let D be a point ta

must fall within the circumference of ken, such that the three

the greater circle. straight lines D A, DB,

Therefore, &c. DC drawn from the

Cor. (Euc. iii. 5 and 6.) If two cirpoint D to the circum

cles cut or touch one another, they canference, are equal to

not have the same centre, one another: the point

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