Let ABC be the given circle, and D the given rectilineal angle; it is required to cut off a segment from the circle ABC that shall contain an angle equal to the angle D.

Draw (III. 17.) the straight line EF touching the circle ABC in the point B, and at the point B in the straight line BF make

(1. 23.) the angle FBC equal

to the angle D.

Then, be

A

E

B

F

cause the straight line EF touches the circle A B C, and BC is drawn from the point of contact B, the angle FBC D is equal (III. 32.) to the angle in the alternate segment BAC of the circle: But the angle FBC is equal to the angle D; therefore the angle in the segment BAC is equal to the angle D: Wherefore the segment BAC is cut off from the given circle ABC containing an angle equal to the given angle D: Which was to be done.

If two straight lines within a circle cut one an.. other, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other.

Let the two straight lines AC, BD, within the circle ABCD, cut one another in the point E: the rectangle contained by AE, EC is equal to the rectangle contained by

BE, ED.

A

If AC, BD pass each of them through the centre, so that E is the centre; it is evident, that AE, R EC, BE, ED, being all equal, the

D

rectangle AE, EC is likewise equal to the rectangle

BE, ED.

D

But let one of them BD pass through the centre, and cut the other AC, which does not pass through the centre, at right angles, in the point E: Then, if BD be bisected in F, F is the centre of the circle ABCD; join AF: And because BD, which passes through the centre, cuts the straight line AC, which does not pass through the centre at right angles in E, AE, EC are equal (III. 3.) to one another: And because the straight line BD is cut into two equal parts in the point F, and into two unequal parts in 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; but the squares of AE, EF are equal (1. 47.) to the square of FA; therefore the rectangle BE, ED, together with the square of EF, is equal to the squares of AE, EF: Take away the common square of EF and the remaining rectangle BE, ED is equal to the remaining square of AE; that is, to the rectangle AE, EC.

A

B

E

Next, let BD, which passes through the centre, cut the other AC, which does not pass through the centre, in E, but not at right angles: Then, as before, if BD be bisected in F, F is the centre of the circle. Join AF, and from F draw (1. 12.) FG per pendicular to AC; therefore AG is equal (III. 3.) to GC; wherefore the rectangle AE, EC, together with the square of EG, is equal (11. 5.) to the square of AG: To each of these squares add the square of GF; therefore the rectangle AE, EC, together

D

A

E

G

B

with the squares of EG, GF, is equal to the squares of AG, GF: But the squares of EG, GF are equal to the square of EF; and the squares of AG, 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 to the rectangle BE, ED together with the square of EF: Take away the common square of EF, and the remaining rectangle AE, EC is therefore equal to the remaining rectangle BE, ED.

IL

Lastly, Let neither of the straight lines AC, BD pass through the centre: Take the centre F, and through E, the intersection of the straight lines AC, DB, draw the diameter GEFH: And because the rectangle AE, EC is equal, as has been shown, to the rectangle GE, EH; and for the A same reason, the rectangle BE, ED is equal to the same rectangle

B

C

GE, EH; therefore the rectangle AE, EC is equal to the rectangle BE, ED. Wherefore, if two straight Q. E. D.

lines, &c.

PROP. XXXVI. THEOR.

If from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square of the line which touches it.

Let D be any point without the circle ABC, and DCA, DB two straight lines drawn from it, of which

DCA cuts the circle, and DB touches the same; The rectangle AD, DC is equal to the square of DB.

D

Either DCA passes through the centre, or it does not; first, let it pass through the centre E, and join EB; therefore the angle EBD is a right angle (III. 18.) And because the straight line AC is bisected in E, and produced to the point D, the rectangle AD, DC, together with the square of EC, is equal (II. 6.) to the square of ED, and CE is equal to EB: B 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 (1. 47.) to the squares of EB, BD, because EBD is a right angle: Therefore the rectangle AD, DC, together with the square of EB, is equal to the squares of EB, BD: Take away the common square of EB; therefore the remaining rectangle AD, DC is equal to the square of the tangent Db.

A

D

E

But if DCA does not pass through the centre of the circle A B C, take (1. 1.) the centre E, and draw EF perpendicular (1. 12.) to AC, and join E B, E C, ED: and because the straight line EF, which passes through the centre, cuts the straight line AC, which does not pass through the centre, at right angles, it shall likewise bisect it (III. 3.); therefore AF is equal to FC: And because the straight line AC is bisected in F, and produced to D, the rectangle

F

A D, DC, together with the square of FC, is equal

(II. 6.) to the square of F D: To each of these equals add the square of F E; therefore the rectangle A D, DC, together with the squares of CE, FE, is equal to the squares of D F, F E But the square of ED is equal (1. 47.) to the squares of D F, F E, because E F D is a right angle; and the square of E C is equal to the squares of CF, FE; therefore the rectangle A D, D C, together with the square of EC, is equal to the square of ED: And CE is equal to EB; therefore the rectangle A D, DC, together with the square of E B, is equal to the square of ED: But the squares of EB, BD are equal to the square (1. 47.) of E D, because E BD is a right angle; therefore the rectangle A D, DC, together with the square of EB, is equal to the squares of E B, BD: Take away the common square of EB; therefore the remaining rectangle A D, DC is equal to the square of D B. Wherefore, if from any point, &c.

COR. If from any point without a circle, there be drawn two straight lines cutting it, as AB, A C, 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 A D which touches the circle.

D

Q. E. D.

B

PROP. XXXVII. THEOR.

A

If from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it; if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, be equal to the square of the line