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PROP. XVIII. THEOR.
If a straight line touch a circle, the straight line drawn from the centre to the point of contact, is perpendicular to the line touching 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 right angle, GCF must be (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 than the greater, which is impossible; wherefore FG is not perpendicular to DE: In the same manner it may be shewn that no other line but FC can be perpendicular to DE; FC is therefore perpendicular to DE. Therefore, if a straight line, &c. Q, E. D.
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 cir cle is 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, FC is perpendicular (13. 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, is the centre; that is, the centre is in CA. Therefore, if a straight line, &c. Q. E. D.
PROP. XX. THEOR.
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 circumfer
Let ABC be a circle, and BDC an angle at the centre and BAC an angle at the circumference, which have the same circumference BC for their base; the angle BDC is double of the angle BAC.
First, let D, the centre of the circle, be within the angle BAC, and join AD, and produce it to E: Because DA is equal to DB, the angle DAB is equal (5. 1.) to the angle DBA; therefore the angles DAB, DBA together are double of the angle DAB; but the angle BDE is equal(32.1.) to the angles DAB, DBA; therefore also the angle BDE is double of the angle DAB: For the same reason, the angle EDC is double of the angle DAC: Therefore the whole angle BDC is double of the whole angle BAC.
Again let D, the centre of the circle, be without the angle BAC; and join AD and produce it to E. It may be demonstrated, as in the first case, that the angle EDC is double of the angle DAC, and that EDB a part of the first, is double of DAB, a part of the other; therefore the remaining angle BDC is double of the remaining angle BAC. There fore the angle at the centre, &c. QE. D
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, both having the same part of the circumference, viz. BCD, for their base; therefore the angle BFD is double (20. c.) of the angle BAD: for the same rea
son, the angle BFD is double of the angle BED: Therefore the an
gle 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 B 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
to the whole angle BED. Wherefore the angles in the same seg. ment. &c. Q. E. D.
PROP. XXII. THEOR.
The opposite angles of any quadrilateral figure described in a circle, aré 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. 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 they 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; and the angles ABC, ADC, are equal to the angles ABC, CAB, BCA. But ABC, CAB, BCA are equal to two right angles (32. 1.); therefore also the angles ABC, ADC are equal to two right angles: In the same manner, the be shewn to be equal to two right angles. angles, &c. QE. D.
angles BAD, DCB may Therefore the opposite
PROP. XXIII. THEOR.
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 circles ACB, ADB, cut one another 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, draw the straight line BCD, and join CA, DA': and because the segment ACB is similar to the segment ADB, and similar segments of circles contain (9. def. 3.) equal angles, the angle ACB is equal to the angle ADB, the exte
rior to the interior, which is impossible (16. 1.). Therefore, there cannot be two similar segments of circles upon the same side of the same line, which do not coincide. Q. E. D.
PROP. XXIV. THEOR.
Similar segments of circles upon equal straight lines are equal to one
Let AEB, CFD be similar segments of circles upon the equal straight lines AB, CD; the segment AEB is equal to the segment CFD. For, if the segment AEB be applied to the segment CFD, so as the point A be on C, and
the straight line AB
upon CD, the point
B shall coincide with
the point D, because
AB is equal to CD: A
Therefore the straight line AB coinciding with CD, the segment AEB must (23. 3.) coincide with the segment CFD, and therefore is equal to it. Wherefore, similar segments, &c. Q. E. D.
PROP. XXV. PROB.
A segment of a circle being given, to describe the circle of which it is the segment.
Let ABC be the given segment of a circle; it is required to describe the circle of which it is the segment.
Bisect (10. 1.) AC in D, and from the point D draw (11. 1.) DB at right angles to AC, and join AB: First, let the angles ÀBD, BAD be equal to one another; then the straight line BD is equal (6. 1.) to DA, and therefore to DC; and because the three straight lines DA, DB, DC, are all equal; D is the centre of the circle (9. 3.): from the centre D, at the distance of any of the three DA, DB, DC, describe a circle ; this shall pass through the other points; and the circle of which ABC is a segment is described and because the centre D is in AC, the segment ABC is a semicircle. Next let the angles ABD, BAD be unequal; at the point A, in the straight line AB make (23. 1.) the angle BAE equal to the angle ABD, and produce BD if neces
sary, to E, and join EC and because the angle ABE is equal to the angle BAE, the straight line BE is equal (6. 1.) to EA: and because AD is equal to DC, and DE common to the triangles ADE, CDE, the two sides AD, DE are equal to the two CD, DE, each to each; and the angle ADE is equal to the angle CDE, for each of them is a right angle; therefore the base AE is equal (4. 1.) to the base EC: but AE was shewn to be equal to EB, wherefore also BE is equal to EC: and the three straight lines AE, EB, EC are therefore equal to one another; wherefore (9. 3.) E is the centre of the circle. From the centre E, at the distance of any of the three AE, EB, EC, describe a circle, this shall pass through the other points; and the circle of which ABC is a segment is described: also, it is evident, that if the angle ABD be greater than the angle BAD, the centre E falls without the segment ABC, which therefore is less than a semicircle; but if the angle ABD be less than BAD, the centre E falls within the segment ABC, which is therefore greater than a semicircle: Wherefore, a segment of a circle being given, the circle is described of which it is à segment. Which was to be done.
PROP. XXVI. THEOR.
In equal circles, equal angles stand upon equal arches, whether they be at the centres or circumferences.
Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their centres, and BAC, EDF at their circumferences: the arch BKC is equal to the arch ELF.
Join BC, EF; and because the circles ABC, DEF are equal, the straight lines drawn from their centres are equal: therefore the two