Page images
PDF
EPUB

+12 Def. 1.

#5. 1.

the circle, as AC; and draw DC to the point C, where it meets the circumference. And because DA is equalt to DC, the angle DAC is equal* to the angle ACD: but DAC is a right angle; therefore ACD is a right angle; and therefore the angles DAC, ACD are equal

+Hyp.

*17. 1.

[merged small][ocr errors][merged small][merged small][merged small]

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 *See fig. 2. without the circle, as AE.*

*12. 1.

*17. 1.

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 be between them: from the point D draw* DG perpendicular 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* +12 Def. than DG: but DA is equal to DH; therefore DH is greater than DG, a part than the whole, which is impossible. Therefore no straight line can be drawn from the point A, between AE and the circumference, which does not cut the circle.

*19. 1.

FE

H

A

COR. From this it is manifest, that the straight line which is drawn at right angles to the diameter of a circle from the t1 Def. 3. 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 *2. 3. is evident, that there can be but one straight line

which touches the circle in the same point.

PROP. XVII. PROB.

To draw a straight line from a given point, either without 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.

*1. 3.

#11. 1

Find the centre E of the circle, and draw AE cutting the circle in D; and from the centre E, at the distance EA, describe the circle AFG; from the point D draw* DF at right angles to EA, and draw EBF, AB. AB shall touch the circle BCD. Because E is the centre of

the circles BCD, AFG, EA is +12 Def. 1. equal to EF, and ED to EB; therefore the two sides AE, EB, are equal to the two FE, ED, each to each; and they contain the angle at E common to the two triangles AEB,

#4. 1.

(CE

FED; therefore the base DF is equal to the base AB, and the triangle FED to the triangle AEB, and the other angles to the other angles: therefore the angle EBA is equal to the angle +Const. EDF: but EDF is a right† angle, wherefore EBA +1 Ax. is a right angle: and EB is drawn from the centre: but a straight line drawn from the extremity of a *Cor. 16. 3. diameter, at right angles to it, touches the circle: therefore AB 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 *Cor. 16. 3. E, and DF at right angles to DE, DF touches the circle.

It is evident that two touching lines may be drawn froin A, as the point F may fall on either side of AE.

[blocks in formation]

If a straight line touch 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 FC: FC shall be perpendicular to DE.

+1. 3.

For, if it be not, from the point F draw FBG perpendicular to DE: and because FGC is a right *17. 1. angle, GCF is* an acute angle; and to the greater *19. 1. angle the greater* side is opposite: therefore FC is +12 Def. 1. greater than FG: but FC is equalt

to FB; therefore FB is greater than FG, the less than the greater, which is impossible: therefore FG is not perpendicular to DE. In the same manner it may be shown, that no other is perpendicular to it besides FC, that is, FC is perpendicular to DE. Therefore, if a straight line, &c. D

Q. E. D.

F

B

GF

This proposition is unnecessary, and may therefore be omitted. It is obviously the converse of the first part of prop. xvi.;—that first part affirming that if a straight line from the extremity of a diameter be perpendicular to that diameter it shall touch the circle, and this 18th proposition states, conversely, that if a straight line touch a circle it will be perpendicular to the diameter from whose extremity it is drawn. But the second part of the 16th is in reality the converse of the first part; since it declares that if a straight line from the extremity of the diameter touch the circle, it can be no other than the perpendicular before mentioned.❤

* The ingenious Williamson, in his excellent edition of the Elements, is the only one, as far as I know, who has adverted to the uselessness of this proposition. He remarks, "There is something rather singular in the 18th proposition, for it seems to me to be nothing but the corollary to the 16th." The corollary, however, is, as Austin justly observes, a superfluous addition; as it is merely a re-statement of the proposition itself in other words. But this, like the other corollaries in the Elements, is supposed to have been subjoined, not by Euclid, but by some one of the early editors.

PROP. XIX. THEOR.

If a straight line touch 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 CA.

For, if not, let F be the centre, if possible, and join CF. Because DE touches the circle ABC, and FC is drawn from the centre to the point of contact, FC is per- B pendicular to DE: therefore FCE is a right angle: but ACE is also a right angle; therefore D

#18.3.

+Hyp.

+1 Ax.

C

E

the angle FCE is equal 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 shown, 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.

This proposition, like the preceding, is quite superfluous, and ought to be expunged from the elements, although we believe it has hitherto escaped all objection. The only thing that it proves is this, viz. that the line which is perpendicular to the diameter must have the diameter perpendicular to it.

PROP. XX. THEOR.

The angle at the centre of a circle is double of the angle at the circumference upon the same arc, 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 arc 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 together double of the angle EAB; but the *32. 1. angle BEF is equal to the angles EAB, EBA; therefore also the

*5. 1.

F

angle BEF is double of the angle EAB: for a like reason the angle FEC is double of the angle EAC: 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.

NOTE. It must not be inferred from this that to every angle at the circumference there corresponds an angle at the centre upon the same arc, and which is double of the former angle. Whenever the angle at the centre really stands upon the same arc, it is necessarily double that at the circumference; but if the angle at the circumference stand upon a semi-circle, then there can be no corresponding angle at the centre; and if the angle at the circumference stand on an arc still greater, then the angle at the centre, corresponding, stands not upon the same, but upon the opposite part of the circumference.

PROP. XXI. THEOR.

The angles in the same segment of a circle are equal to one another.

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

« PreviousContinue »