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BC, FG, and join EB, EC, EF: and because AE is equal + to EB, and ED to EC, therefore AD is equal+ + 2 Ax.

+ 15 Def. 1. to EB, EC: but EB, EC are greater

* than BC; where- * 20. 1. fore also AD is greater than BC.

And, because BC is nearer + to the centre than FG, + Hyp. EH is less * han EK: but, as was demonstrated in the • 5 Def. 3. preceding, BC is double of BH, and FG double of FK, and the squares of EH, HB are equal to the squares EK, KF: but the square of EH is less than the square of EK, because EH is less than EK; therefore the square of BH is greater than the square of FK, and the straight line BH greater than FK, and therefore BC is greater than FG.

Next, let BC be greater than FG; BC shall be nearer to the centre than FG, that is + the same construction + 5 Def. 3. being made, EH shall be less than EK. Because BC is

greater than FG, BH likewise is greater than KF: and the squares of BH, HE are equal to the squares of FK, KE; of which the square of BH is greater than the square of FK, because BH is greater than FK: therefore the square of EH is less than the square of EK, and the straight line EH less than EK: and therefore BC is nearer + to the centre than FG. Wherefore + 5 Def. 3. the diameter, &c.

Q. E. D.

PROP. XVI. THEOR. The straight line drawn at right angles to the diameter of See N.

a circle, from the extremity of it, falls without the circle ; and no straight line can be drawn from the extremity, between that straight line and the circumference, so as not to cut the circle ; or, which is the same thing, no straight line can make so great an acute angle with the diameter at its extremity, or so small an angle with the straight line which is at right angles to it, as not to cut the circle.

Let ABC be a circle, the centre of which is D, and the diameter AB: the straight line drawn at right angles to AB from its extremity A, shall fall without the circle.

For, if it does not, let it fall, if possible, within the circle, as AC; and draw DC to the point C, where it meets the

A circumference. And because DA is equalt to DC, the angle DAC is equal*

+15 Def. 1.

B В

* 5. 1.

* 17.1.

• 17. 1.

* 19. 1.

+ Hyp. to the angle ACD: but DAC is a right+ angle; there

fore ACD is a right angle; and therefore the angles DAC, ACD are equal to two right angles; which is * impossible: therefore the straight line drawn from A at right angles to BA does not fall within the circle. In the same manner it may be demonstrated, that it does not fall upon the circumference; therefore it must fall without the circle, as AE.

Also, between the straight line AE and the circumference no straight line can be drawn from the point A

which does not cut the circle. For, if possible, let AF * 12. 1. be between them: from the point D draw* DG per

pendicular to AF, and let it meet the circumference in H. And because AGD is a right angle, and DAG less*

than a right angle, DA is greater than DG: but DA +15 Def. 1. is equal † to DH; therefore DH is greater than DG, the less than the

FE greater, which is impossible: therefore no straight line can be drawn from the

H point A between AE and the circum B В ference, which does not cut the circle : or, which amounts to the same thing, however great an acute angle a straight line makes with the diameter at the point X, or however small an angle it makes with AE, the circumference must pass between that straight line and the perpendicular AE. “ And this is all that is to be understood, when, in the Greek text, and translations from it, the angle of the semicircle is said to be greater than any acute rectilineal angle, and the remaining angle less than any rectilineal angle.” Q. E. D.

Cor. From this it is manifest, that the straight line

which is drawn at right angles to the diameter of a cir+ 2 Def. 3. cle from the extremity of it, touchest the circle; and

that it touches it only in one point, because, if it did meet the circle in two, it would fall * within it. “ Also it is evident, that there can be but one straight line which touches the circle in the same point.”

A

2. 3.

PROP. XVII. PROB.

To draw a straight line from a given point, either with

out or in the circumference, which shall touch a given circle.

First, let A be a given point without the given circle

BCD ; it is required to draw a straight line from A which shall touch the circle.

Find * the centre E of the circle, and join AE; and • 1. 3. from the centre E, at the distance EA, describe the circle AFG; from the point D draw * DF at right an- * 11. 1. gles to EA, and join EBF, AB. AB shall touch the circle BCD.

Because E is the centre of the circles BCD, AFG, EA is equalt to

+ 15 Def. 1. EF, and ED to EB; therefore the two sides AE, EB, are equal to the

GICE two FE, ED, each to each; and they contain the angle

at E common to the two triangles AEB, FED; therefore the base DF is equal* to the base AB, and the triangle FED to the • 4. 1. triangle AEB, and the other angles to the other angles: therefore the angle EBA is equal to the angle EDF: but EDF is a right+ angle, wherefore EBA is a right+ I Const.

+1 angle; and EB is drawn from the centre: but a straight line drawn from the extremity of a diameter, at right angles to it, touches * the circle: therefore AB .Cor. 16.3. touches the circle; and it is drawn from the given point A. Which was to be done.

But if the given point be in the circumference of the circle, as the point D, draw DE to the centre E, and DF at right angles to DE: DF touches * the circle.

* Cor. 16.3.

PROP. XVIII. THEOR.

If a straight line touches a circle, the straight line drawn

from the centre to the point of contact shall be perpendicular to the line touching the circle.

Let the straight line DE touch the circle ABC in the point C; taket the centre F, and draw the straight line

† 1, 3. FC: FC shall be perpendicular to DE.

For, if it be not, from the point F draw FBG perpendicular to DE: and because FGC is a right angle, GCF is *

an acute angle; and to the greater angle the • 17. 1. greater * side is opposite: therefore FC is greater than FG: but FC is equal to

+ 15 Def. 1. FB ; therefore FB is greater than FG, the less than the greater, which is impos

F sible: therefore FG is not perpendicular to DE. In the same manner it may

be

C GE shewn, that no other is perpendicular to

* 19.1.

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it besides FC, that is, FC is perpendicular to DE. Therefore, if a straight line, &c. Q. E. D.

PROP. XIX. THEOR.

If a straight line touches a circle and from the point of

contact a straight line be drawn at right angles to the touching line, the centre of the circle shall be in that line.

Let the straight line DE touch the circle ABC in C, and from C let CA be drawn at right angles to DE: the centre of the circle shall be in CĂ.

For, if not, let F be the centre, if possible, and join CF. Because DE touches the circle ABC, and FC is drawn from

BI the centre to the point of contact, FC is perpendicular* to DE; therefore FCE

D

E is a right angle: but ACE is also a right+ angle; therefore the angle FCE is equalt to the angle ACE, the less to the greater, which is impossible: therefore F is not the centre of the circle ABC. In the same manner it

may be shewn, that no other point which is not in CA, is the centre; that is, the centre is in CA. Therefore, if a straight line, &c. Q. E. D.

F

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PROP. XX. THEOR.

See N.

The angle at the centre of a circle is double of the angle

at the circumference upon the same base, that is, upon the same part of the circumference.

Let ABC be a circle, and BEC an angle at the centre, and BAC an angle at the circumference, which have the same circumference BC for their base : the angle BEC shall be double of the angle BAC.

Join AE, and produce it to F. First, let the centre of the circle be within the angle BAC. Because EA is equal to EB, the angle EAB is equal* to the angle EBA; therefore the angles EAB, EBA are double of the angle EXB; but the angle BEF is equal * to the

B angles EAB, EBA; therefore also the angle BEF is double of the angle EAB: for the same reason the angle FEC is double of the angle EAC:

* 5. 1.

32. 1.

А

E

therefore the whole angle BEC is double of the whole angle BAC.

Again, let the centre of the circle be without the angle BAC. It may be demonstrated, as in the first case, that the angle FEC is double of the angle FAC, and that FEB, a part of the first, is double of FAB, a part of the other; therefore the remaining angle BEC is double of the remaining angle BAC. Therefore the angle at the centre, &c. Q. E. D.

F

B

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PROP. XXI. THEOR. The angles in the same segment of a circle are equal to See N.

one another. Let ABCD be a circle, and BAD, BED angles in the same segment BAED: the angles BAD, BED shall be equal to one another.

B В First, let the segment BAED be greater than a semicircle. Take + F,

† 1. 3. the centre of the circle ABCD, and join BF, FD: and because the angle BFD is at the centre, and the angle BAD at the circumference, and that they have the same part of the circumference, viz. BCD for their base; therefore the angle BFD is double* of the angle

* 20. 3. BAD: for the same reason the angle BFD is double of the angle BED: therefore the angle BAD is equalt + 7 Ax. to the angle BED.

Next, let the segment BAED be not greater than a semicircle. Draw AF to the centre, and produce it to C, and join CE: therefore the segment BADC is greater than a semicircle; and the angles in it BAC, BEC are equal, by the first case: for the same reason, because CBED is greater than a semicircle, the angles CAD, CED, are equal: therefore the whole angle BAD is equal+ + 2 Ax. to the whole angle BED. Wherefore the angles in the same segment, &c. Q. E. D.

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F

PROP. XXII. THEOR.

The opposite angles of any quadrilateral figure inscribed

in a circle, are together equal to two right angles.

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