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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 rectilineal 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.

C

H

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 line

A

F

B

AB, make (23. 1.) the angle BAD equal

AD; bisect (10. 1.) AB in F, and

to the angle C, and from the point A draw (11. 1.) AE at right angles to

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 equal (4. 1.) to the base GB; and the circle described from the centre G, at the distance GA, shall pass

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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. 1. 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 describ

H

C

F

B

A

G

E

D

ed which contains an angle equal to the given angle at C.

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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.

A

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 angle FBC is equal to the angle D therefore the angle in the segment BAC is equal to the angle

D

E

B

F

D: wherefore the segment BAC is cut off from the given circle ABC containing an angle equal to the given angle D.

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 tc 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 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 cen- B

A

E

D

tre, 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

C

the circle ABCD; join AF: and because BD, which passes through the

centre, cuts the straight line AC, which does not

D

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.) + EF2 = FB2 = AF2.
AF2 · A E2 + (47. 1.) EF2, therefore BE.ED +
EF2, = AE2 + EF2, and taking EF2 from each,
BE.ED AE2-AE.EC.

=

But

Next, let BD, which passes through the centre, cut the other AC, which does not pass through

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D

the centre, in E, but not at right angles; then, as before, if BD be bisect. ed in F, F is 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.) EG2 AG2, and adding GF2 to both, AE.EC+EG2+GF2=AG2+GF2. Now EG2+GF2=EF2, and AG2+GF2=AF2; therefore AE.EC+EF2=AF2=FB2. But FB2 A =BE.ED+(5. 2.) EF2, therefore AE.EC+ EF2 =BE.ED+EF2, 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.

F

E

G

C

B

H

F

E

A

C

B G

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.

D

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. 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 (18. 3.) and because the straight line AC is bisected in E, and produced to the point D, AD.DC+EC2=ED2 (6. 2.). But EC EB, therefore AD.DC + EB2 = ED2. Now ED2 (47. 1.) EB2+ BD2, because EBD

=

is a right angle; therefore AD.DC + EB2 : EB2 + BD2, and taking EB2 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

B

C

E

A

the straight line AC, which does not pass through the centre, at right angles, it likewise bisects it (3. 3.); therefore AF is equal to FC; and because the straight line AC is bisected in F, and produced to D (6. 2.), AD.DC+FC2= FD2; add FE2 to both, then AD.DC+FC2+ FE2=FD2+FE2. But (47. 1.) EC2=FC2+ FE2, and ED2-FD2+FE2, because DFE is a right angle; therefore AD.DC+EC2-ED2. Now, because EBD is a right angle, ED2= EB2+BD2 EC2+BD2, and therefore, AD. DC+EC2=EC2+BD2, and AD.DC=BD2.

COR. 1. 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.

COR. 2. It follows, moreover, that two tangents drawn from the same point are equal.

COR. 3. And since a radius drawn to the point of contact is perpendicular to the tangent, it follows that the angle included by two tangents, drawn from the same point, is bisected by a line drawn from the centre of the circle to that point; for this line forms the hypotenuse common to two equal right angled triangles.

B

F

D

E

A

A

D

PROP. XXXVII. THEOR.

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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.

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 angle (18. 3.): 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 DF 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 diameter, and the straight line which is drawn at right angles to a diameter, from the extremity of it, touches (16. 3.) the circle therefore DB touches the circle ABC.

:

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ADDITIONAL PROPOSITIONS.

PROP. A. THEOR.

A diameter divides a circle and its circumference into two equal parts; and, conversely, the line which divides the circle into two equal parts is a diameter

Let AB be a diameter of the circle AEBD, then AEB, ADB are equal in surface and boundary.

Now, if the figure AEB be applied to the figure ADB, their common base AB retaining its position, the curve line AEB must fall on the curve line ADB ; otherwise there would, in the one or the other, be points unequally distant from the centre, which is contrary to the definition of a circle.

A

E

F

B

D

Conversely. The line dividing the circle into two equal parts is a diameter

For, let AB divide the circle into two equal parts; then, if the centre is not in AB, let AF be drawn through it, which is therefore a diameter, and consequently divides the circle into two equal parts; hence the portion AEF is equal to the portion AEFB, which is absurd.

COR. The arc of a circle whose chord is a diameter, is a semicircumference, and the included segment is a semicircle.

PROP. B. THEOR.

Through three given points which are not in the same straight line, one circumference of a circle may be made to pass, and but one.

Let A, B, C, be three points not in the same straight line: they shall all lie in the same circumference of a circle.

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