been proved equal to BAD: therefore the remaining angle DBE is equal to the angle BCD in the alternate segment of the circle. Wherefore, if a straight line, &c. Q. E. D.

PROP. XXXIII. PROB.

Upon a given straight line to describe a segment of a circle, containing an angle equal to a given rectilineal angle.

Let AB be the given straight line, and the angle at C the given rec tilineal angle; it is required to describe upon the given straight line AB a segment of a circle, containing an angle equal to the angle C. First, let the angle at C be a right angle; bisect (10. 1.) AB in F, and from the centre F, at the distance FB, describe the semicircle AHB; the angle AHB being in a semicircle is (31. 3.) equal to the right angle at C.

But if the angle C be not a right

angle at the point A, in the straight

A

F

H

B

line AB, make (23.1.) the angle BAD equal to the angle C, and from

the point A draw (11. 1.) AE

at right angles to AD; bisect (10. 1.) AB in F, and from F draw (11. 1.) FG at right angles to AB, and join GB: Then because AF is equal to FB, and FG common to the triangles AFG, BFG, the two sides AF, FG are equal to the two BF, FG; but the angle AFG is also equal to the angle BFG; therefore the base AG is e

H

H

E

F

B

B

qual (4. 1.) to the base GB; and the circle described from the centre G, at the distance GA, shall pass through the point B; let this be the circle AHB: And because from the point A the extremity of the diameter AE, AD is drawn at right angles to AE, therefore AD (Cor. 16. 3.) touches the circle; and because AB, drawn from the point of contact A, cuts the circle, 'the angle DAB is equal to the angle in the alternate segment AHB (32. 3.); but the angle DAB is equal to the angle C, therefore also the angle C is equal to the angle in the segment AHB:

Wherefore, upon the given straight line AB the segment AHB of a circle is described which contains an angle equal to the given angle at C. Which was to be done.

PROP. XXXIV. PROB.

To cut off a segment from a given circle which shall contain an angle equal to a given rectilineal angle.

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 (17.3.) the straight line EF touching the circle ABC in the point B, and at the point B, in the straight line BF make (23. 1.) the angle FBC equal to the angle D; therefore, because the straight line EF touches the circle ABC, and BC is drawn from the point of contact B, the angle FBC is equal (32. 3.) to the angle in the alternate segment BAC: but the D angle FBC is equal to the angle D; therefore the angle in the

E

B

F

C

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.

PROP. XXXV. THEOR.

If two straight lines within a circle cut one another, 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.

If AC, BD pass each of them through the centre, so that E is the centre, it is evident, A that AE, EC, BE, ED, being all equal, the rectangle AE.EC is likewise equal to the rectangle BE.ED.

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

B

D

E

point E: then, if BD be bisected in F, F is the centre of the circle

D

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 (3. 3.) to one another: and because the straight line BD is cut into two equal parts in the point F, and into two unequal, in the point E, BE.ED (5. 2.)+ EF FB = AF2. But AF2 = AE+. (47. 1.) EF2, therefore BE.ED+EF= AE2+EF2, and taking EF2 from each, BE. ED-AE AE.EC.

F

A

E

D

F

B

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 betore, if BD be bisected in F, Fis the centre of the circle. Join AF, and from F draw (12.1.) FG perpendicular to AC; therefore AG is equal (3. 3) to GC; wherefore AE.EC+(5. 2.) EG-AG2, and adding GF to both, AÉ.EC+EG +GF”=AG3 +GF9. Now EGGF EF, and AG+GF-AF2; therefore AE.EC+EF-AF-FB.

=

A

B

But FB2

=BE.ED+(5.2.) EF2,therefore AE.EC+EF2=BE.ED+EF, and taking EF2 from both, AE.EC.-BE.ED.

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 as has been shown, AE.EC=GE.EH, and BE.ED= GE.EH; therefore AE.EC =BE.ED. A Wherefore, if two straight lines, Q. E. D.

c.

H

D

F

C

十

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, is 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 it: the rectangle AD.DC is equal to the square of DB.

M

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 (18. 3.) angle: and because the straight line AC is bisected in E, and prodaced to the point D, AD.DC + EC2 ED (6. 2.). But EC-EB, therefore AD DC+EB2 ED2. Now ED (47. 1.) EB2+BD2, because EBD is a right angle; therefore AD.DC+EB2=EB + BD2, and taking EB from each, AD.DC=BD2.

But, if DCA does not pass through the centre of the circle ABC, take (1. 3.) the centre E, and draw EF perpendicular (12. 1.) to AC, and join EB, EC, 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 likewise besects (3.,3.) it; therefore AF is equal to FC; and because the straight line AC is bisected in F, and produced to D (6. 2.), AD.DC+FC= FD2; add FE2 to both, then AD.DC+FC2 +FE FD2+FE. But (47. 1.) EC-FC9

FE, and ED2 FD2+ FE, because DFE is a right angle; therefore AD.D+ EC-ED. Now, because EBD is a right angle, ED-EB +BD-EC2+BD, and therefore, AD.DC+EC EC2+BD, and AD.DC-BD".

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, AC, the rectangles contained by the whole lines and the parts of them without the circle, are equal to one another, viz. BA.AE CA.AF; for each of these rectangles is equal to the square of the straight line AD, which touches the circle.

E

PROP. XXXVII. THEOR.

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 which meets it, the line which meets shall touch the circle.

Let any point D be taken without the circle ABC, and from it let two straight lines DCA and DB be drawn, of which DCA cuts the circle, and DB meets it; if the rectangle AD.DC, be equal to the square of DB, DB touches the circle.

D

Draw (17. 3.) the straight line DE touching the circle ABC; find the centre F, and join FE, FB, FD; then FED is a right (18. 3.) angle: and because DE touches the circle ABC, and DCA cuts it, the rectangle AD.DC is equal (36. 3.) to the square of DE; but the rectangle AD,DC is, by hypothesis, equal to the square of DB: therefore the square of DE is equal to the square of DB; and the straight line DE equal to the straight line DB: but FE is equal to FB, wherefore DE, EF are equal to DB, BF; and the base FD is common to the two triangles DEF. DBF; therefore the angle DEF is equal (8. 1.) to the angle DBF; and DEF is a right angle, therefore also DBF is a right angle; but FB, if produced, is a B diameter, and the straight line which is drawn at right angles to a diameter, from the extre mity of it, touches (16. 3.) the circle: therefore DB touches the circle ABC. Wherefore, if from a point, &c. Q, E. D.