Book III. PROP. XIX. THEOR. F a ftraight line touches a circle, and from the point of contact a ftraight line be drawn at right angles to the touching line, the centre of the circle is 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 centre of the circle is in CA. For, if not, let F be the centre, if poffible, and join CF: Because DE touches the circle no other point which is not in CA, is the centre; that is, the centre is in CA. Therefore, if a ftraight line, &c. Q. E. D. THE PROP. XX. THEOR. HE angle at the centre 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 ABC be a circle, and BDC an angle at the centre, and BAC an angle at the circumference, which have the fame circumference BC for their bafe; the angle BDC is double of the angle BAC. First, ict D, the centre of the circle, be within the angle BAC, and join AD, and produce it to E: Becaufe DA is equal equal to DB, the angle DAB is E D A Again let D, the centre of the circle, be without the angle BAC, and join AD and produce it to E. It may be demonftrated, as in the first cafe, 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. Therefore the angle at the centre, &c. Q. E. D. E B PROP. XXI. THEOR. THE angles in the fame fegment of a circle are equal 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 centre of the circle ABCD And, firft, let the fegment BAED be greater than a femicircle, and join BF, FD: And because the angle BFD is at the centre, and the angle B F D BAD N. Book III. BAD at the circumference, and that they have the fame part of the circumference, viz. BCD, for their bafe; therefore the a 20. 3. angle BFD is double a of the angle BAD: for the fame reafon, the angle BFD is donble of the angle BED: Therefore the angle BAD is equal to the angle BĚD. But, if the fegment BAED be not greater than a femicircle, let BAD, BED be angles in it; these allo are equal to one BA E D F C fore the whole angle BAD is equal to the whole angle BED. Wherefore the angles in the fame fegment, &c. Q. E. D. TH PROP. XXII, THEOR. THE oppofite angles of any quadrilateral figure defcribed in a circle, are together equal to two right angles. Let ABCD be a quadrilateral figure in the circle ABCD; any two of its oppofite angles are together equal to two right. angles. Join AC, BD. The angle a 21. 3. CAB is equal a to the angle CDB, because they are in the D A B each each of these equals add the angle ABC; and the angles Book III. ABC, ADC, are equal to the angles ABC, CAB, BČA. But ABC, CAB, BCA are equal to two right angles; there- b 32. I. fore alfo the angles ABC, ADC are equal to two right angles: In the fame manner, the angles BAD, DCB may be fhewn to be equal to two right angles. Therefore the oppofite angles, &c. Q. E. D. UPO PON the fame ftraight line, and upon the fame fide of it, there cannot be two fimilar fegments of circles, not coinciding with one another. If it be poffible, let the two fimilar segments of circles, viz. ACB, ADB, be upon the fame fide of the fame 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 D cut one another in any other point a : one of the fegments must therefore fall within the other: let ACB fall within ADB, draw the ftraight line A BCD, and join CA, DA: and be B a 10. 3 cause the fegment ACB is fimilar to the fegment ADB, and fimilar fegments of circles contain b equal angles, the b 9. def. 3. 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 circles upon the fame fide of the fame line, which do not coincide. Q. E. D. PROP. Book III. S IMILAR fegments of circles upon equal ftraight lines, are equal to one another. Let AEB, CFD be fimilar fegments of circles upon the equal straight lines AB, CD; the fegment AEB is equal to the fegment CFD. For, if the fegment AEB be applied to the fegment CFD, fo as the A point A be on C, and the ftraight line AB upon CD, the point B fhall coincide with the point D, becaufe AB is equal to CD: Therefore the ftraight line AB coinciding with CD, the fegment AER a 23. 3. muft coincide with the fegment CFD, and therefore is equal to it. Wherefore, fimilar fegments, &c. Q. E. D. a 10. I. c 6. 1. A PROP. XXV. PROB. SEGMENT of a circle being given, to defcribe the circle of which it is the fegment. Let ABC be the given fegment of a circle; it is required to defcribe the circle of which it is the fegment. C Bifecta AC in D, and from the point D drawb DB at b II. I. right angles to AC, and join AB: First, let the angles ABD, BAD be equal to one another; then the ftraight line BD is equal to DA, and therefore to DC; and because the three ftraight lines DA, DB, DC, are all equal; D is the centre 9. 3. of the circled: from the centre D, at the diftance of any of the three DA, DB, DC, defcribe a circle; this fhall pafs through the other points; and the circle of which ABC is a fegment is defcribed and because the centre D is in AC, the fegment ABC is a femicircle. But if the angles ABD, BAD are not equal to one another, at the point A, in the ftraight |