At the point A, in the straight line AB, make the angle BAD equal to the angle C (I. 23); and from the point A draw AE at

right angles (I. 11) to AD. Eisect AB in F (I. 10); and from F draw FG at right angles (I. 11) to AB. Join GB.

Because AF is equal to FB, and FG common to the triangles AFG and BFG, the two sides AF and FG are equal to the two sides BF and FG, each to each. But the angle AFG is equal (I. Def. 10) to the angle BFG. Therefore the base AG is equal (I. 4) to the base GB. With centre G, and distance GA, describe the circle AHB, and it shall pass through the point B. The segment AHB contains an angle equal to the given rectilineal angle C.

Because from the point A, the extremity of the diameter AE, the straight line AD is drawn at right angles to AE, the straight line AD touches the circle (III. 16). Because AB is drawn from the point of contact A, cutting the circle, the angle DAB is equal (III. 32) to the angle in the alternate segment AHB. But the angle DAB is equal (Const.) to the angle C. Therefore the angle C is equal (I. Ax. 1) to the angle in the segment AH B. Wherefore, upon the given straight line AB, the segment of a circle AHB is described, which contains an angle equal to the given angle C. Q. E. F.

PROP. XXXIV. (PROBLEM.)-From a given circle (ABC) to cut off a segment, which shall contain an angle equal to a given rectilineal angle (D).

Draw the straight line EF touching the circle ABC in the point B (III. 17), and at the point B, in the straight line BF, make the angle FBC equal (I. 23) to the angle D. segment BAC contains an angle equal to the given angle D.

The

E

A

B

Because the straight line EF touches the circle ABC, and BC is drawn from the point of contact B, the angle FBC is equal (III. 32) to the angle in the alternate segment BAC of the circle. But the angle FBC is equal (Const.) to the angle D. Therefore the angle in the segment BAC is equal (I. Ax. 1) to the angle D. Wherefore, from the given circle ABC, the segment BAC is cut off, containing an angle equal to the given angle D. Q. E. F.

PROP. XXXV. (THEOREM.)-If in a circle (ABCD) two straight lines (AC, BD) cut one another, the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other.

First, let AC and BD pass each through he centre E. Because the segments AE, EC, BE and ED are all equal (I. Def. 15).

Therefore the rectangle AE. EC, is equal to the rectangle BD.ED.

Secondly, let the one BD pass through the centre, and cut the other AC, which does not pass through the centre, at right angles, in the point E.

B

D

E

Find F, the centre of the circle ABCD, and join AF. Because BD passing through the centre, cuts the straight line AC, which does not pass through the centre, at right angles in E (III. 3), AE is equal to EC. Because the straight line BD is cut into two equal parts at the point F, and into two unequal parts at the point E, the rectangle BE.ED, together with the square of EF, is equal (II. 5) to the square of FB; that is, to the square of FA, because FA is equal to FB. But the squares of AE and EF, are equal (I. 47) to the square of FA. Therefore the rectangle BE.ED together with the square of EF, is equal (I. Acc. 1) to the squares of AE and EF. From these equals take away the common square of EF. Therefore the remaining rectangle BEED is equal (I. Ax. 3) to the remaining square of AE; that is, to the rectangle AE. EC.

A

E

B

Thirdly, let BD, passing through the centre, cut the other AC, which does not pass through the centre, at E, but not at right angles. Find F, the centre of the circle, join AF, and from F draw FG perpendicular (I. 12) to AC.

D

E

G

B

Because AG is equal (III. 3), to GC, the rectangle A E. EC, together with the square of EG, is equal (II. 5) to the square of AG. To each of these equals add the square of GF. Therefore the rectangle AE. EC, together with the squares of EG and GF, is equal (I. Ax. 1) to the squares of AG and GF. But the squares of EG and GF are equal (I. 47) to the square of EF. And the squares of AG and GF are equal to the square of AF. Therefore the rectangle AE. EC, together with the square of EF, is equal to the square of AF; that is, to the square of FB. But the square of FB is equal (II. 5) to the rectangle BE ED, together with the square of EF. Therefore the rectangle AE. EC, together with the square of EF, is equal (I. Ax. 1) to the rectangle BE. ED, together with the square of EF. From these equals take away the common square of EF. Therefore the remaining rectangle AE. EC, is equal (4x. 3) to the remaining rectangle BE. ED. Lastly, let neither of the straight lines AC and BD cutting each other, pass through the centre. Find the centre F (III. 1), and through E, the point of intersection of the straight lines AC and DB, draw the diameter GH.

Because the rectangle AE. EC is equal, by the preceding case, to the rectangle GE.EH; and the rectangle BE. ED is equal to the same rectangle GE.EH; therefore the rectangle AE. EC is equal (I. Ax. 1) to the rectangle BE. ED. Wherefore, if two straight lines, &c. Q. E. D.

H

B

PROP. XXXVI. (THEOREM.)—If from any point (D) without a circle (ABC) two straight lines (DA, DB) be drawn, one of which cuts the circle, and the other touches it; the rectangle contained by the whole line (DA) which cuts the circle, and the part of it without the circle, is equal to the square of the tangent (DB, DC).

First, let DA pass through the centre E.

Join EB, and (III. 18) EBD is a right angle.

C

Because the straight line AC is bisected in E, and produced to D, the rectangle 'AD.DC. together with the square of EC, is equal (II. 6) to the square of ED. But CE is equal to EB. Therefore the rectangle AD.DC, together with the square of EB, is equal to the square of ED. But the square of ED is equal (I. 47) to the squares of EB and BD. Therefore the rectangle B AD.DC, together with the square of EB, is equal (I. Ax. 1) to the squares of EB and BD. From these equals take away the common square of EB. Therefore the remaining rectangle AD.DC is equal (Ax. 3) to the square of the tangent DB.

A

Next, let DA not pass through the centre of the circle ABC. Find E, the centre of the circle (III. 1), draw EF perpendicular to AC (I. 12) and join EB, EC, and ED.

F

Because the straight line EF, passing through the centre, cuts the straight line AC, which does not pass through the centre, at right angles, it bisects AC (III. 3); therefore AF is equal to FC. Because the straight line AC is bisected in F, and produced to D, the rectangle AD.DC, toge with the square of FC, is equal (II. 6) to the square of FD. To each of these equals add the square of FE. Therefore the rectangle AD. DC, together with B the squares of CF and FE is equal (I. Ax. 2) to the squares of DF and FE. But the square of ED is equal (I. 47) to the squares of DF and FE. Álso the square of EC is equal to the squares of CF and FE. Therefore the rectangle AD.DC, together with the square of EC, is equal (I. Ax. 1) to the square of ED. But CE is equal to EB. Therefore the rectangle AD.DC, together with the square of EB, is equal to the square of ED. But the squares of EB and BD are equal (I. 47) to the square of ED. Therefore the rectangle AD. DC, together with the square of EB, is equal to the squares of EB and BD. From these equals take away the common square of EB. Therefore the remaining rectangle AD.DC is equal (1. Ax. 3) to the square of DB. Wherefore, if from any point, &c. Q. E. D.

Cor.-If from any point without a circle, there be drawn two straight lines cutting it, as AB and AC, the rectangles contained by the whole lines and the parts of them without the circle, are equal to one another; viz., the rectangle BA.AE, to the rectangle CA.AF: for each of them is equal to the square of the straight line AD, which touches the circle.

A.

B

PROP. XXXVII. (THEOREM.)-If from a point (D) without a circle (ABC) there be drawn two straight lines (DA, DB), one of which (DA) cuts the circle, and the other (DB) meets it; and if the rectangle (AD. DC) contained by the whole line which cuts the circle, and the part of it without the circle, be equal to the square of (DB) the straight line which meets it, that straight line touches the circle.

Draw the straight line DE, touching the circle ABC, in the point E (III. 17). Find F, the centre of the circle (III. 1); and join FE, FB, and FD.

D

Because FED is a right angle (III. 18), DE touches the circle ABC. But DA cuts the circle (Hyp.) Therefore the rectangle AD. DC is equal (III. 36) to the square of DE. But the rectangle AD.DC is (Hyp.) equal to the square of DB. Therefore the square of DE is equal (I. Ac. 1) to the square of DB, and DE to DB. But FE is equal to FB (I Def. 15). Wherefore the two sides DE and EF are equal to the two sides DB and BF, ach to each; and the base FD is common to the two triangles DEF and DBF. Therefore the angle DEF is equal (I. 8) to the angle DBF. But DEF is a right angle. Therefore, also, DBF is a right angle (4x. 1), and BF, if produced, is a diameter. But the straight line passing through the extremity of a diameter, at right angles to it (III. 16), touches the circle. Therefore DB touches the circle ABC. Wherefore, if from a point, &c. Q. E. D.

A

ONE rectilineal figure is said to be inscribed in another, when all the angular points of the inscribed figure are upon the sides of the figure in which it is inscribed, each upon each.

II.

In like manner, one rectilineal figure is said to be described about another, when all the sides of the circumscribed figure pass through the angular points of the figure about which it is described, each through

each.

III.

A rectilineal figure is said to be inscribed in a circle, when all the angular points of the inscribed figure are upon the circumference of the circle.

IV.

A rectilineal figure is said to be described about a circle, when each side of the circumscribed figure touches the circumference of the circle.

V.

In like manner, a circle is said to be inscribed in a rectilineal figure, when the circumference of the circle touches each side of the figure.

VI.

A circle is said to be described about a rectilineal figure, when the circumference of the circle passes through all the angular points of the figure about which it is described.

VII.

A straight line is said to be placed in a circle, when the extremities of it are in the circumference of the circle.

PROP. I. (PROBLEM.)-In a given circle (ABC) to place a straight line, equal to a given straight line (D) which is not greater than the diameter. Find the centre of the circle ABC (III. 1), and draw the diameter BC. If BC is equal to D, what is required is done; that is, in the circle ABC, a straight line BC is placed equal to D. But if BC is not equal to D, it is greater than D (Hyp.) From CB cut off CE equal to D (I. 3). From C as centre, at the distance CE, describe the circle AEF; and join CA.

Because C is the centre of the circle D AEF, CA is equal (I. Def. 15) to CE.

But CE is equal (Const.) to D. Therefore CA is equal (I. Ax. 1) to D. Wherefore in the circle ABC, a straight line CA is placed equal to the given straight line D, which is not greater than the diameter of the circle. Q. E. F.

PROP. II. (PROBLEM.)-In a given circle (ABC) to inscribe a triangle equiangular to a given triangle (DEF).

Draw the straight line GH touching the circle in the point A (III. 17). At the point A, in the straight line AH, make the angle HAC equal (1. 23) to the angle DEF. At the point A, in the straight line AG, make the angle GAB equal to the angle DFE. Join BC; the triangle ABC is the triangle required.

E

C

D

H

Because GH touches the circle ABC, and AC is a chord drawn from the point of contact, the angle HAC is equal (III. 32) to the angle ABC in the alternate segment of the circle. But the angle HAC is equal to (Const.) the angle DEF. Therefore also the angle ABC is equal (I. Ax. 1) to DEF. For the same reason, the angle ACB is equal to the angle DFE. Therefore the remaining ångle BAC is equal (1. 32 and Ax. 1) to the remaining angle EDF. Wherefore the triangle ABC is equiangular to the triangle DEF, and it is inscribed in the circle ABC. Q. E. F.

PROP. III. (PROBLEM.)-About a given circle (ABC) to describe a triangle (DEF) equiangular to a given triangle.

Produce BF both ways to the points G and H. Find the centre K