it is required to draw a straight line from A which fhall touch Book III. the circle. Find the center E of the circle, and join AE; and from a 1. 3. the center E, at the diftance EA, defcribe the circle AFG; from the point D draw DF at right angles to EA, and join b 11. 1. EBF, AB; AB touches the circle BCD. Because E is the center of the circles BCD, AFG, EA is equal to EF: And ED to EB; therefore the two fides AE, EB are equal to the two FE, ED, and they contain the angle at E common to the two triangles AFB, FED; therefore the bafe DF is equal to the bafe AB, and the triangle FED to the triangle AEB, and the other angles to the other angles: Therefore the c 4. I. 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 center; but a ftraight line drawn from the extremity of a diameter, at right angles to it, touches the circle: Therefore d Cor.16.3. 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 center E, and DF at right angles to DE; DF touches the circle. IF a straight line touches a circle, the straight line drawn from the center to the point of contact, fhall be perpendicular to the line touching the circle. Let the ftraight line DE touch the circle ABC in the point C, take the center F, and draw the ftraight 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 is an acute b 17. I. angle; and to the greater angle the greateft fide is oppofite: © 19. I. Therefore F 2 C Book III. Therefore FC is greater than FG; but FC is equal to FB; therefore FB is greater than FG, the lefs than the greater, which is impoffible: Wherefore FG is not perpendicular to DE: In the fame manner it may be fhewn, that no other is perpendicular to it befides FC, that is, FC is perpendicular to DE. Therefore, if a ftraight line, &c. Q. E. D. B D C GE a 18.3. IF F a ftraight line touches a circle, and from the point of contact a straight line be drawn at right angles to the touching line, the center of the circle fhall be in that line. Let the ftraight line DE touch the circle ABC in C, and from C let CA be drawn at right angles to DE; the center of the circle is in CA. A F For, if not, let F be the center, if poffible, and join CF: Because DE touches the circle ABC, and FC is drawn from the center to the point of contact, FC is perpendicular 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 lefs to B the greater, which is impoffible: Wherefore F is not the center of the circle ABC: In the fame manner, it may be fhewn, that no other point D C E See N. that is, the center is in CA. Therefore, if a straight line, &c. Q. E. D. THE HE angle at the center of a circle is double of the angle at the circumference, upon the fame bafe, that is, upon the fame part of the circumference. Let Let ABC be a circle, and BEC an angle at the center, and Book III. BAC an angle at the circumference, which have the fame circumference BC for their bafe; the angle BEC is double of the angle BAC. a First, Let E the center of the circle bé within the angle BAC, and join AE, and produce it to F: 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 EAB; but the angle BEF is equal to the angles B EAB, EBA; therefore alfo the angle BEF is double of the angle EAB: For the fame reafon, the angle FEC is double of b A the angle EAC: Therefore the whole angle BEC is double of the whole angle BAC. Again, Let E the center of the circle be without the angle BDC, and join DE and produce it to G. It may be demonftrated, as in the firft cafe, that 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 re- G maining angle BEC is double of the remaining angle BDC. Therefore the angle at the center, &c. Q. E. D. B E PROP. XXI. THEOR. THE angles in HE angles in the fame fegment of a circle are e- See N. qual to one another. Let ABCD be a circle, and BAD, BED angles in the fame fegment BAED: The angles BAD, BED are equal to one another. Take F the center of the circle ABCD: And, firft, let the fegment BAED be greater than a femicircle, and join BF, FD: And because the B angle BFD is at the center, and the angle BAD at the circumference, and that they have the fame part of E F 3 the Book III. the circumference, viz. BCD, for their base; therefore the an ngle BFD is double of the angle BAD: For the fame reafon, the angle BFD is double of the angle BED: Therefore the angle BAD is equal to the angle BED. a 20.3. 2 32. 1. b 21. 3. B A E D But, if the fegment BAED be not greater than a femicircle, let BAD, BED be angles in it; thefe alfo are equal to one another: Draw AF to the center, and produce it to C, and join CE: Therefore the fegment BADC is greater than a semicircle; and the angles in it BAC, BEC are equal, by the firft cafe: For the fame reason, because CBED is greater than a femicircle, the angles CAD, CED are equal: Therefore the whole angle BAD is equal to the F C whole angle BED. Wherefore the angles in the fame segment, &c. Q. E. D. HE oppofite angles of any quadrilateral figure defcribed in a circle, are together equal to two right THE angles. Let ABCD be a quadrilateral figure in the circle ABCD; any two of its oppofite angles are together equal to two right angles. a Join AC, BD; and because the three angles of every triangle are equal 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 to the angle CDB, because D B qual to the angles ABC, ADC: But ABC, CAB, BCA are equal to two right angles; therefore alfo the angles ABC, ADC are equal to two right angles: In the fame manner, the angles BAD, BAD, DCB may be fhewn to be equal to two right angles. Book. III. Therefore, the oppofite angles, &c. Q. E. D. PON the fame ftraight line, and upon the fame fide see N. of it, there cannot be two fimilar fegments of UPON circles, not coinciding with one another. If it be poffible, let the two fimilar fegments of circles, viz. ACB, ADB, be upon the fame fide of the fame ftraight 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: One of the fegments muft therefore fall within the other; let ACB fall within ADB, and draw the ftraight line BCD, and join CA, DA: And because the segment ACB is fimilar to the fegment A B a Io. 3. ADB, and that fimilar fegments of circles contain b equal an- b 11. def. 3. gles; the angle ACB is equal to the angle ADB, the exterior to the interior, which is impoffible. Therefore, there cannot c 16. 1. be two fimilar fegments of a circle upon the same side of the fame line, which do not coincide. Q. E. D. IMILAR fegments of circles upon equal ftraight See N. lines, are equal to one another. Let AEB, CFD be fimilar fegments of circles upon the equal ftraight lines AB, CD; the fegment AEB is equal to the fegment CFD. For, if the feg ment AEB be E B C the straight line D AB upon CD, the point B fhall coincide with the point D, be E 4 caufe |