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angles of equiangular triangles. Also, because the triangles KB A, EDC are similar, B K is to KA as DE to E C; and for the like reason, KA is to K L as EC to EF: therefore, ex æquali, BK: KL::DE: EF. And in the same manner it may be demonstrated, that the sides about the other equal angles of the two figures are proportionals. Therefore, the figure described upon A B is similar to the given figure CDEFG. (def. 14.)

Method 2. Place A B parallel to CD, and join CA, D B. Then, if AB be not equal to CD, CA will not be parallel to D B (I. 21. and 14. Cor. 2.), and CA, DB will meet, if produced, in some point P: join P E, P F, PG: through B draw BK parallel to DE (I. 48.), and let it meet PE in K: through K draw K L parallel to EF, and let it meet PF in L: through L draw L M parallel to F G, and let it meet PG in M; and join AM: the figure ABKLM shall be similar to the given figure CDEFG.

G

K

For, because PC: PA:: PD: PB (29.), that is, ::PE: PK, that is :: PF: PL, that is, :: PG: P M, AM is parallel to CG; and because the sides of the two figures are parallel, each to each, they are equiangular (I. 18.) with one another. Also, because the triangles CDP, A BP are similar, CD: DP::AB: B P, and for the like reason DP: DE: BP BK; therefore, ex @quali, CD: DE::AB: BK. And in the same manner it may be demonstrated, that the sides about the other equal angles of the two figures are proportionals. Therefore the figure upon A B is similar to the given figure. When AB is equal to CD, CA is parallel to B D, and the lines EK, FL, GM, instead of being drawn to a point P, must be drawn parallel to C A or B D. The demonstration that the figure ABKLM, so constructed, will be both similar and

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To describe a rectilineal figure, which shall be similar to a given rectilineal figure CDEFG, and shall have its perimeter equal to a given straight line AQ.

Produce CD to R, (see the second may be figure of Prop. 65.) so that CR equal to the perimeter of the given figure CDEFG; take A B (53.) a fourth proportional to CR, AQ, and CD, and upon AB (65.) describe a rectilineal figure AB KLM similar to the given figure. Then, because the figures AB KLM and CDEFG are similar, the perimeter of the first is to the perimeter of the other (43.), as AB to CD, that is (12.), as A Q to CR: but the perimeter of CDEFG is equal to CR: therefore (18. Cor.) the perimeter of AB KLM 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 rectilineal figure DEF.

B

K

M

E

Find (I. 58. Cor.) the straight lines G, H such that their squares may be equal to the figures ABC, DEF respectively: take KL a fourth proportional (53.) to G, H, and B C, and upon KL (65.) describe the figure KLM similar to the figure BČA. Then, because the figure A B C is to the figure MK L as the square of BC to the square of K L (43.), that is, (37. Cor. 4.) as the square of G to the square of H, and that the figure A B C is equal to the square of G, the figure K L M is (18.) equal to the square of H, that is, to the figure DE F: and it is similar to ABC; therefore it is the figure required.

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Describe (I. 54.) the triangle ABG; equal to the figure ABCDEF, and naving the side AB, and angle A B C the same with it: divide (55.) the base BG in the given ratio in the point H: join AC and AD: through H draw HK parallel to AC, to meet CD produced in K; through K draw KL раrallel to AD to meet D E in L, and join AL: AL shall be the line required.

For, by the construction, which is similar to that of I. 55., it is evident, that the figure ABCDL is equal to the triangle AB H: but the whole figure is equal to the triangle AB G: therefore, the part ALE F is equal to the triangle AHG. Therefore, the parts of the figure are as the triangles A B H, AHG, that is (39.) as the bases BH, HG, that is, in the given ratio.

Next, let P be the given point in the side A B, from which the line of division is to be drawn.

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Describe, as before, (I. 54.), the triangle A B G equal to the given figure, and, by drawing AG' parallel to PG, make the triangle PB G'equal to ABG (as in I. 56.): divide (55.) B G' in the given ratio in the point H: join PC and PD: through H draw HK parallel to P C, to meet CD produced in K: through K draw K L parallel to PD to meet DE in L, and join PL. Then,

• The line AH is wanting in this figure; and PH is in like manner wanting in the figure below,

as also AG, AG'.

as before, the part PBCDL of the given figure, is equal to the triangle PBH, and the remaining part PAFEL to the triangle PH G'. Therefore, the parts of the figure are as the bases BH, HG' (39.), that is, in the given ratio. Therefore, &c.

PROP. 69. Prob. 19.

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 CE for one of its angles, and the opposite side passing through the point B.

have three cases to consider; first, when In the solution of this problem we the point B is in one of the given lines without the given angle DCE; and as CD; secondly, when the point B is thirdly, when B is within the angle DCE.

CD. Take CD equal to a side of the triCase 1. Let the point B be in the line angle A, and upon CD (I. 50.) describe a triangle CDF, which shall have its two remaining sides equal to the two remain. ing sides of the triangle A, each to each, and therefore (I. 7.) shall be equal to the

B

T

triangle A in every respect: through F (I. 48.) draw F E parallel to DC: join DE, BE: through D (I. 48.) draw DG 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 E C D, that is (because E F is parallel to CD) to F C D, that is, to the given triangle A.

Case 2. Let the

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point B be without the angle DCE. Through B (I. 48.) draw BD parallel to CE to meet CD in D, and by the construction pointed out in Case 1. describe a triangle DCF having the given side D C and the given angle DCE, and equal to the given triangle A: in CF produced take the

point G such that CGx GF may be equal to BDXC F (56.): join BG, and let BG cut C D in H: CHG shall be the triangle required.

Join HF. Then, because CGX GF is equal to BD×CF, CF: FG::CG : BD (38.), that is, since DHB and CHG are (I. 15.) equiangular, ::CH : HD. Therefore (29.) H F is parallel to DG; and hence, as before, the triangles DHF and GHF are (I. 27) equal to one another, and CHG is equal to CDF, that is to A.

Case 3. Let the point B be within the angle D C E. Through B (I. 48.) draw BD parallel to EC to meet CD in D, and, by the construction pointed out in Case 1, describe

D

H

a triangle D C F, having the given side CD and the given angle DCE, and equal to the given triangle A ; in CF (if it be possible) take the point G such that CGXGF may be equal to BDXCF (56. Cor.): join BG, and let GB produced cut CD in H; CHG shall be the triangle required.

which shall form
with two given
straight lines CD
CE cutting one
another in Ca
triangle equal to
a given triangle
A;" we should
have had four
solutions, two
of them corre-
sponding to the
angle in which
B lies, and une

CHD h

D'

for each of the adjacent angles; as is apparent from the foregoing constructions applied to the adjoined figure. When B D is equal to a fourth of CF, two of these become identical; when BD exceeds a fourth of CF, the same two become impossible, but the other two, viz. those which correspond to the adjacent angles, are always possible, except when B is in one of the lines as CD. This last-mentioned position is peculiar: it has likewise, however, two solutions, one for each of the adjoining angles.

Join HF. Then, by a demonstration § which may be given in the same words as that of Case 2, the triangle C H G is equal to CD F, that is to the given triangle A.

In this last case a solution will be impossible if BD exceed a fourth of CF; for then BDXCF will exceed a fourth of the square of CF, that is (I. 29. Cor. 2.) the square of half CF; and no point G can be taken in CF such that CG x GF may exceed the square of half C F (56. N. B.).

We may remark also that, whenever the solution is possible in the last case, two points G may be found such that CG × GF BD x CF, (56.) and therefore two lines G H may be drawn satisfying the given conditions. When BD is exactly a fourth of CF, these two points coincide with one another and with the middle point of CF, and therefore the two solutions become identical.

If B be in one of the lines DC, EC produced, or within the angle which is vertical to DCF, the solution will be manifestly impossible.

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BOOK III.

1. First Properties of the Circle-
$2. Of Angles in a Circle-§ 3. Rect-
angles under the segments of Chords
-4. Regular Polygons, and Ap-
proximation to Circular Area—§ 5.
Circle a Maximum of Area, and
Minimum of Perimeter-§ 6. Simple
and Plane Loci-§ 7. Problems.

SECTION 1.-First Properties of the
Circle.

Def. 1. Any portion of the circum-
ference of a circle is called an erc; and
the straight line which
joins the extremities of
an arc is called the chord
of that arc.

When the chord passes through the centre, it is a diameter, and the arcs upon either side of it, being equal to one another, are called, each of them, a semi-circumference.*

2. The figure which is contained by an arc and its chord is called a segment.

That every diameter divides a circle and its cir cumference into two equal parts, is evident from the symmetrical character of the circle. The same may be proved by doubling the figure upon the diameter; for, every point of the circumference being at the same distance from the centre, the parts so applied

(whether of circumference or area) will coincide

A segment which is contained by a semicircumference and a diameter, is equal to half the circle, and is therefore called

a semicircle.

3. An angle in a segment is the angle contained by two straight lines drawn from any point of the arc of the segment to the extremities of its 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.

By this is intended no more than that when the term "equal circles" may be hereafter used, those which have equal radii are to be understood. That such circles, however, have also equal areas, is at once evident by applying one to the other, so that their centres may

coincide.

6. Concentric circles are those which have the same centre.

7. (Euc. iii. def. 2.) A straight line is said to touch a circle, when it meets the circumference in any point, but being produced does not cut it in that point. Such a line is frequently, for brevity's sake, called a tangent; and the point in which it meets the circumference is called the point of contact.

8. Circles are said to touch one another, when they meet, but do not cut one another.

9. A rectilineal figure is said to be inscribed in a circle, when all its angular points are in the circumference of the circle. Also, when this is the case, the circle is said to be circumscribed about the rectilineal figure.

PROP. 1. (EUc. iii. 2.)

ference of a circle in two points, it shall If a straight line meet the circumout the circle in those points; and the part of the straight line which is between them shall fall within the circle.

Let the straight line AB meet the circumference of a circle having the

centre C in the points A and B: it shall cut the circle in those points, and the part AB shall fall within the

circle.

join CD: take also the points E and F Join CA, CB: bisect A B in D, and in the line A B, the former between A B, and join CE, CF. and B, the latter upon the other side of

Then, because CA is equal to CB, CAB is an isosceles triangle; therefore (I. 6. Cor. 3.) C D, which is drawn from the vertex to the bisection of the base, is perpendicular to A B. And, because CE is nearer to the perpendicular than CB is, (I. 12. Cor. 2.) CE is less than CB; therefore the point E is within the circle: on the other hand, because C F is farther from the perpendicular than C B is, CF is greater than CB; and therefore the point F is without the circle. And the same may be said of the parts about A. Therefore the straight line in question cuts the circle in the points A and B.

Also, because every point, as E, of A B, is at a less distance from C than the radius, the part AB falls within

the circle.

Therefore, &c.

Cor. 1. A straight line cannot meet a circle in more than two points.

Cor. 2. A straight line which touches a circle meets it in one point only. Cor. 3. A circle is concave towards

10. A rectilineal figure is said to be its centre. circumscribed about

a circle, when all its sides touch the circle. Also, when this is the case, the circle is said to be inscribed in the rectilineal figure.

PROP. 2. (EUC. iii. 16.)

The straight line, which is drawn at right angles to the radius of a circle from its extremity, touches the circle; and no other straight line can touch it in the same point,

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For, CA being shorter (I. 12. Cor. 3.) than any other line which can be drawn from C to A T, every other point or AT lies without the circle: therefore AT meets the circle in the point A but does not cut it, that is, it touches the circle.

In the next place, let DE be any other straight line passing through the same point A: DE shall cut the circle. For, CA not being perpendicular to DE, let CE be perpendicular to it. Then, because CE is less than CA (I. 12. Cor. 3.), the point E is within the circle. But if D be a point in DE upon the other side of A, C D will be farther from the perpendicular than CA; wherefore CD being (I. 12. Cor. 2.) greater than CA, the point D will be without the circle. Therefore the straight line DE cuts the circle; and the same may be demonstrated of every other straight line which passes through A, except the straight line AT only, which is at right angles to A C.

Therefore, &c.

Cor. 1. (Euc. iii. 18.) If a straight line touches a circle, the straight line drawn from the centre to the point of contact, shall be perpendicular to the line touching the circle.

Cor. 2. (Euc. iii. 19.) If a straight line touches a circle, and from the point of contact, a straight line be drawn at right angles to the touching line, the centre of the circle shall be in that line.

Cor. 3. Tangents TA, TB which are drawn to a circle from the same point T, are equal to one another. For, CAT, CBT being right-angled triangles which have the common hypotenuse CT, and their sides CA, CB equal to one another, their remaining sides TA, TB are likewise equal. (I. 13.) Cor. 4. Tangents which are at the extremities of the same diameter are parallel to one another.

PROP. 3. (EUC. iii. 3.)

If a diameter cut any other chord at right angles, it shall bisect it; and

conversely, if a diameter bisect any other chord, it shall cut it at right angles.

Let AB be a diameter of the circle ADE, the centre of

which is C; and,

first, let it cut the chord DE at right angles in the point F: DE shall be bisected in F.

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Join CD, CE. Then, because CDE is an isosceles triangle, the straight line CF which is drawn from the vertex C at right angles to the base D E, bisects the base, that is, DF is equal to FE. (I. 6. Cor. 3.)

Next let D E be bisected in F by the diameter A B; the angles D FA, É FA shall be right angles. For in the isosceles triangle CDE, the straight line which is drawn from the vertex C to the middle point of the base DE is at right angles to the base. (I. 6. Cor. 3.)

Therefore, &c.

Cor. 1. A diameter bisects all chords which are parallel to the tangent at either extremity of the diameter.

Cor. 2. The straight line which bisects any chord at right angles passes through the centre of the circle.

Cor. 3. If two circles have a common chord, the straight line which bisects it at right angles shall pass through the centres of both the circles.

Cor. 4. (Euc. iii. 4.) It appears from the proposition, that two chords of a circle cannot bisect one another except they both of them pass through the centre.

For, if only one of the chords pass through the centre, it cannot be bisected by the other which does not pass through the centre.

And if, when neither of them passes through the centre, it were possible that they should bisect one another, the diameter passing through the supposed point of mutual bisection would be at right angles to each of them, which is absurd.

PROP. 4. (EUc. iii. 15.)

The diameter is the greatest straight line in a circle; and, of others, that which is nearer to the centre is greater than the more remote: also the greater is nearer to the centre than the less.

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