2. Let C be any oblique angle (d): make, at the point A, angles BAD, equal to C, and DAE a right angle (e); draw FG at right angles to AB (f); join GB.

Now, on account of the right angles at F, and that FA, FG are equal to FB, FG, the bases GA, GB are equal (g); and about G as centre, a circle may be drawn to pass through the points A, E, B, of the greater segment, and A, H, B, of the less; and DA shall touch the circle in A (h). Therefore the angle AEB is equal to BAD, or C, acute; and the angle AHB is equal to BAD, or C, obtuse.

Therefore, upon the given straight line, a segment has been described, containing an angle given in variety of magnitude, as right, acute, and obtuse.

Recite (a) p. 10, 1;
(d) Note def. 8, 1;

(b) p. 31, 3;

(c) ax. 10, 1;

(e) p. 23, 1;

(f) p. 11, 1:

(g) p. 4, 1, and cor. p. 1, 3;

(h) p. 31, 32, and cor.

p. 16, 3.

34 P. To cut off a segment from a given circle (ABC), which shall contain an angle equal to a given rectilineal angle (D).

Constr. Draw any tangent, as EF, touching the given circle in a point B (a); and at the point B, in the straight line FB, make the angle FBC equal to the angle D (b).

Argument. Because EF touches the circle; and from B, the point of contact, a chord BC is drawn in the circle; there are two angles equal to the two in the alternate segments (c): therefore the angles FBC

E

and BAC are equal; but FBC was made equal to the given angle D; and so, BAC is equal to D (d); and it is in a segment cut off from a given circle; which was to be done.

35 Th. If two chords (AC, BD) in a circle cut each other, the rectangle of the segments (AE, EC) of one of the chords, is equal to the rectangle of the segments (BC, CD) of the other.

1. If the sectional point E, be the centre of the circle; then it is obvious that AEXEC equals BEX ED: for the segments are equal radii (a).

A

2. If the chords cut each other at right angles in E, but not in the centre, one of them AC is bisect- B ed in E (b), the other passes through the centre F, where it is bisected (c): join AF. Now, since BD is divided equally in F, and unequally in E, BF2, or AF BEX ED+EF2 (d)=AE2+ EF2 (e): from each of these equals take EF2; then BEXED=AE2, AEXEC: for AE equals EC.

3. If BD pass through the centre F, and cut AC obliquely in E, draw FG perpendicular to AC (b); join AF: therefore, since BD is bisected in F, and AC in G, and both unequally divided in E,

BF2=BEXED+EF2(d); and EF2=EG2+GF2(e);

AG

AEXEC+EG2(d); to these add GF2 (ƒ); then, AG2+GF2-AEXEC+EG2+GF2=AF2(e)= BF2: therefore, BEXED+EG2+GF2=AEXEC+ D EG2+GF; from which equals taking EG+GF2, which are common, BEXED=AEXEC (ƒ).

4. If neither AC nor BD pass through the centre F, draw the diameter GEFH; then since either of the rectangles AEXEC, or BEX ED proves equal to GEXEH, as above; in this case also AEXEC= BEXED (ƒ).

Wherefore, if two chords in a circle, &c.

Recite (a) def. 15, 1; (d) p. 5, 2;

H

Q. E. D. (b) p. 3, and cor. p. 1,3; (c) def. 14, 1; (e) p. 47, 1; (f) ax. 1, 2, 3.

36 Th. If from any point (D) without a circle (ABC) two straight lines be drawn, one of which (DCA) cuts, and the other (DB) touches the circle, the rectangle of the secant and its exterior part is equal to the square of the tangent.

1. If the secant DCA, pass through the centre of the circle E, join EB: then EBD is a right angle (a); and DE2 DB2+BE (b); and because AC is bisected in B E and produced to D, DE-AD×DC+CE2 (c); therefore DB2+BEAD×DC+CE2 (d): take now from both sides the squares of the radii BE, CE, the remainders are equal (e); namely, DB2=AD×DC.

2. If the secant DCA pass on one side of the centre E; join EB, EC, ED, and draw EF perpendicular to AD (ƒ); therefore DE2=DF2+FE2 (b)=DB2+BE2, as above and because AC is bisected in F and produced to D, DF2=AD+DC CF2(c); to both sides add FE2; therefore DF2+FE2=AD×DC+CF2+FE2 (g): but CF2+FE2=CE2 (b), or BE2; substituting the latter, DF2+ FE2=ADXDC+BE2; therefore DB2+BE2= ADXDC+BE: take from these BE2, which is common, DB2=ADXDC (e),

Cor. If from any point A, without a circle, two secants AEB, AFC be drawn; the rectangle of the one AB, and its exterior part AE, is equal to the rectangle of the other AC, and its exterior part AF: for each of these rectangles is equal to the square of the tangent AD (d).

Wherefore, if from any point without, &c.

B

Q. E. D.

B

Recite (a) p. 18, 3;
(d) ax. 1;
(g) ax. 2.

(b) p. 47, 1;
(e) ax. 3;

(c) p. 6, 2; (f) p. 12, 1; also p. 3, 3;

37 Th. If from a point (D), without a circle (ABC), two straight lines be drawn, one (DB) meeting, and the other (DEA) cutting the circle; if the rectangle of the cutting line and its exterior part (DC), be equal to the square of the line which meets the circle, the latter line shall touch and not cut the circle.

Draw the tangent DE (a); find the centre of the circle F (b); join FE, FD, FB: then FED is a right angle (c) and because DE touches and DCA cuts the circle, DE ADX DC (d); but DB2 is given equal to the same rectangle; therefore DB2 equals DE, and DE is equal to DB; the radii FB, FE are also equal, and FD is common to the two triangles DBF, DEF; which have therefore three sides in the one equal to Α

three sides in the other; therefore (e) the angle DBF is equal to the angle DEF; and the latter being a right angle by construction, the former is also a right angle; and so, the straight line DB touches the circle (f).

Therefore, if from a point without a circle, &c.

38 P. To construct a rectangle equal to a given square, having the difference of its adjacent sides equal to a given line.

Let C be the side of a given square, and AB the difference of the sides of an equivalent rectangle. Describe a circle on the diameter AB; at the extremity A, of the diameter, draw the tangent AD, equal to C; through the centre O, draw DF, B cutting the circle in E, F. Then the rectangle DEX DF equals the square of AD (a), or its equal C; and the difference of DE and DF is EF, which is equal to AB.

Recite (a) p. 36 of b. 3

A

BOOK FOURTH.

Definitions.

1. One rectilineal figure is said to be inscribed in another, when all the angles of the former are upon the sides of the latter.

Note.-Regular polygons, which have the same number of equal sides, thus inscribed, form a series, and have a certain ratio to each other.

2. One rectilineal figure is described about another, when all the sides of the former pass through the angular points of the latter.

3. A rectilineal figure is inscribed in a circle, when the angles of the former are all in the circumference of the latter. And in this case, the circle is said to be described about the rectilineal figure.

4. A rectilineal figure is described about a cir

cle, when each side of the former touches the circumference of the latter. And in this case, the circle is said to be inscribed in the rectilineal figure.

5. A straight line is said to be placed in a circle, when its extreme points are in the circumference.

Propositions.

1 P. In a given circle (ABC) to place a chord equal to a given straight line (D), not greater than the diameter.

Draw BC, the diameter of the circle: then if the given straight line D, be equal to BC, there is placed in the circle such a chord as was required.

But if D be not equal to BC, it cannot be greater (a); make CE equal to D (b); and with CE as radius, upon the centre C, describe the circle AEF (c); join CA (d): therefore, in the circle AEF, CA and CE are

equal radii (e): but CE is equal D; therefore CA is equal to D; and it is a chord (f), not greater than the diameter, placed in the circle ABC; which was to be done.

Recite (a) p. 15, 3;

(d) pos. 1;

(b) p. 3, 1;
(e) def. 15, 1;

(c) pos. 3;
(f) def 18 1.