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that AB is double of AF, and CD double of CG, and that the squares of EF, FA are equal to the squares of EG, GC; of which the square of FE is equal to the square of EG, because FE is equal to EG; therefore the remaining square of AF is equal to the remaining square of CG; and the straight line AF is therefore equal to CG: And AB is double of AF, and CD double of CG: wherefore AB is equal to CD. Therefore equal straight lines, &c. Q. E. D.

Proposition Xy. Theorem. The diameter is the greatest straight line in a circle : and, of all others, that which is nearer to the centre is always greater than one more remote ; and the greater is nearer to the centre ihan the less.

Let ABCD be a circle of which the diameter is AD, and the centre
E; and let BC be nearer to the centre than FG;
AD is greater than any straight line BC which is
not a diameter, and BC greater than FG.
From the centre draw EH, EK perpendiculars

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to BG, FC, and join EB, EC, EF: and because
AE is equal to EB, and ED to EC, AD is equal to
EB, EC: But EB, EC are greater (20. 1.) than
BC; wherefore, also, AD is greater than BC.

And, because BC is nearer to the centre than FG, EH is less (5. Def. 3.) than EK : But, as was demonstrated in the preceding, BC is double of BH, and FG double of FK, and the squares of EH, HB are equal to the squares of EK, KF, of which 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 is nearer to the centre than FG, that is, the same construction being made, EH is less than EK : Because BC is greater than FG, BH likewise is greater than FK: 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. Wherefore the diameter, &c. Q. E. D.

Proposition. XVI. Theorem. The straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls without the circle, and no straight line can be drawn between that straight line and the circumference from the extremity, 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 circie.

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

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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 circumference : And because DA is equal to DC, the angle DAC is equal (5.1.) BI to the angle ACD; but DAC is a right angle, therefore ACD is a right angle, and the angles DAC, ACD are therefore equal to two right angles ; which is impossible. (17. 1.) 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.

And 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 FA be between them, and from the point D draw (12. 1.) DG perpendicular to FA, and let it meet the circumference in H: And because AGD is a right angle, and DAG less (19. 1.) than a right angle: DA is greater (19. 1.) than DG : But DA is equal to DH ; therefore DH is greater than DG, the less to the greater, which is impossible: Therefore no

FE straight line can be drawn from the point A between AE and the circuinference, 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 A, or, however small an angle it makes with AE, the circumference passes 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.'

Cor. From this it is manifest, that the straight line which is drawn at right angles to the diameter of a circle from the extremity of it, touches 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. (2. 3.) • Also it is evident that there can be but one straight line which touches the circle in the same point.

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Proposition. XVII. Problem.

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.

Find (1.3.) the centre E of the circle, and join AE; and from the centre E, at the distance EA, describe the circle AFG; from the point D draw (11.1.) DP at right angles to EA, and join EBF, AB. AB touches the circle BCD

Because E is the centre of the circles BCD, AFG, EA is equal to EF: And ED to EB; therefore the two sides AE, EB are equal to the two FE, ED, and they contain an angle at E com. mon to the two triangles AEB, 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 : (4. 1.) Therefore the angle EBA is equal to the angle EDF:

But EDF is a right angle, wherefore EBA is a right angle: And EB is drawn from the centre : But a straight line drawo from the extremity of a diameter, at right angles to it, touches the circle : (Cor. 16.3.) Therefore AB touches the circle ; and it is

3 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 io the centre E, and DF at right angles to DE; DF touches the circle.

Proposition XVIII. Theorem. 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 louching the circle.

Let the straight line DE touch the circle ABC in the point C; take the centre F, and draw the straight line FC: FC is perpendicular to

DE.

For, if it be not, from the point F draw FBG perpendicular to DE; and because FGC is a righi angle, GCF is (17. 1.) an acute angle; and to the greater angle the greater (19. 1.) side is opposite : Therefore FC is greater than FG; but FC is equal to FB; therefore FB is greater than FG the less to the greater, which is impossible: Wherefore FG is not perpendicular to DE: In the same manner it may be shewn, that no other is perpendicular to it besides FC, that is, FC is perpendicular to DE. Therefore, if a straight line, &c. Q. E. D.

Proposition XIX. Theorem. 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 is 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, PC is perpendicular (18.3.) to DE; therefore FCE is a right angle : But ACE is also a right angle; therefore the angle FCE is equal to the angle ACE the less to the greater, which is impossible : Wherefore 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,

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is the centre ; that is, the centre is in CA. Therefore, if a straight line, &c. Q. E. D.

Proposition XX. Theorem.

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 is double of the angle BAC.

First, let E, the centre of the circle, be within the angle BAC, and join AE and produce it to F: Because EA is equal to EB, the angle EAB is equal (5. 1.) to the angle EBA ; therefore the angles EAB, EBA are double of the angle EAB; but the

B angle BEF is equal (32. 1.) to the 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 EXC. Therefore the whole angle BEC is double of the whole angle BAC.

Again, let E the centre of the circle, be without the angle BDC, and join DE, and produce it to G. It may be demonstrated, as in the first case, that

D

E the angle GEC is double of the angle GDC, and that GEB a part of the first, is double of GDB, a part of the other ; therefore the remaining angle BEC is double of the remaining angle BDC. Therefore the angle at the centre, &c. Q. E. D.

B

Proposition XXI. Theorem.

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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: The angles BAD, BED are equal to one another.

Take F, the centre of the circle ABCD: And first, let the segment BAED be greater than a semicircle, 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 (20. 3.) of the angle BAD: For the same reason, the angle BFD is double of the angle BED; Therefore the angle BAD is equal to the angle BED.

But if the segment BAED be not greater than a semicircle, let BAD, BED be angles in it; these, also, are equal to one another : Draw AF to the centre, and produce it to C, and join CE: Tbere

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fore tae 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 semi-circle, the angles CAD, CED F
are equal : Therefore the whole angle BAD is equa.
to the whole angle BED. Wherefore the angles in
he same segment, &c. Q. E. D.

Proposition XXII. Theorem. The opposite angles of any quadrilateral figure described in a circle, are together equal to two right angles.

Let ABCD be a quadrilateral figure in the circle ABCD; any two of its opposite angles are together equal to two right angles.

Join AC, BD, and because the three angles of every triangle are equal (32. 1.) to two right angles, the three angles of the triangle CAB. viz. the angles CAB, ABC, BCA, are equal to two right angles : But the angle CAB is equal (21. 3.) to the angle CDB, because they are in the same segment BADC, and the angle ACB is equal to the angle ADB, because there are in the same segment ADCB : Therefore the whole angle ADC is equal to the angles CAB, ACB : To each of these equals add the angle ABC; therefore the angles ABC, CAB, BCA, are equal to the angles ABC, ADC: But ABC, CAB, BCA are equal to two right angles ; therefore also the angles ABC, ADC, are equal to two right angles : In the same manner the angles DAB, DCB may be shewn to be equal to two right angles. Therefore, the opposite angles, &c. Q. E. D.

D

Proposition XXIII. Theorem. Upon the same straight line, and upon the same side of it, there cannot be two similar segments of circles, not coinciding with one another.

If it be possible, let the two similar segments of circles, viz. ACB, ADB, be upon the same side of the same straight line AB, not coinciding with one another : Then, because the circle ACB cuts the circle ADB in the two points A, B, they cannot cut one another in any other point : (10. 3.) One of the segments must, therefore, fall within the other : Let ACB, fall within ADB, and draw the straight line BCD, and join CA, DA: And because the segment ACB is similar to the segment ADB, and that similar segments of circles contain (11. Def.3.) equal angles; the angle ACB is equal to the angle ADB, the exterior to the interior opposite, which is impossible. (16. 1.) Therefore, there cannot be two similar segments of a circle upon the same side of the same line, which do not coincide Q E. D.

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