angles of equiangular triangles. Also, because the triangles K BA, 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 DB (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 KL 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.

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 BD, and the lines EK, FL, GM, instead of being drawn to a point P, must be drawn parallel to CA or B D. The demonstration that the figure ABKLM, so constructed, will be both similar and

PROP. 66. Prob. 16.

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.

be

Produce CD to R, (see the second figure of Prop. 65.) so that CR may 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.)

D

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. Find (1.58. 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 K L M similar to the figure BČA. Then, because the figure A B C is to the figure MKL 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 DEF: and it is similar to ABC; therefore it is the figure required,

K

M

EE

H

point G such that CGx GF may be equal to BDxCF (56.): join BG, and let BG cut CD 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 DCE. Through B (I. 48.) draw BD parallel to EC to meet CD in D, and, by the construction pointed out in Case 1, describe 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 BDX CF (56. Cor.): join BG, and let GB produced cut CD in H; CHG shall be the triangle required.

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 CDF, that is to the given triangle A.

E

cl

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 one
for each of the adjacent angles; as is
apparent from the foregoing construc-
tions applied to the adjoined figure.
When B D is equal to a fourth of C F,
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, ex-
cept 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.

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 = BDXC F, (56.) and therefore two lines G H may be drawn satisfying the given conditions. When B D 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.

Scholium.

Had the problem been proposed under the following form: " through a given point B to draw a straight line

CHD

D'

BOOK III.

$1. First Properties of the Circle-
§ 2. Of Angles in a Circle-§ 3. Rect-
angles under the segments of Chords
-84. Regular Polygons, and Ap-
proximation to Circular Area-§ 3.
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 arc; 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

H

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 Isaid to touch one another, when they meet, but do not cut one another.

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

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 C B, 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 AB. 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, C F 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.

9. A rectilineal figure

the circle.

Also, because every point, as E, of 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 AB falls within 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.

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. 1. (EUC. iii. 2.)

If a straight line meet the circumference of a circle in two points, it shall out 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

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

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.