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 chat line. * 18. 3. 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 circles is in CA. For, if not, let F be the centre, if possible, and join CF: because DE touches the circle A F С E 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. PROP. XX. THEOR. See Note. THE angle at the centre of a circle is double of the a 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 Book 111. BAC an'angle at the circumference, which have the same circumference BC for their base; the angle A First, Let E the centre of the circle be E a 5. 1. angle EBA; therefore the angles EAB, FBA are double of the angle EAB; but the angle BEF is, equal to the angles b 32... EAB, EBA; therefore also the angle BEF с A D E a PROP. XXI. THEOR. THE angles in the same segment of a circle are See Note: equal to one another. : Book 111. the circumference, viz. BCD, for their base; therefore the an no gle BFD) is double a of the angle BAD: for the same reason, a 20.3. 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 А. E D F С 40th L. PROP. XXII. THEOR. THE opposite angles of any quadrilateral figure described in a circle, are together equal to two right angles. a 52. 1. b 21. 3. 1 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 thrce angles of every triangle are equal a to two right angles, the three angles of the triangle ÇAB, viz. the angles CAB, ABC, BCA are equal to two right angles: but the angle CAB D B : BAD, DCB may be shown to be equal to two right angles. Book III. 1 PROP. XXIII. TIIEOR. UPON the same straight line, and upon the same See Note: side of it, there cannot be two similar segments of circles, not coinciding with one another. : a 10. 3. If it be possible, let the two similar segments of circles, viz. D B с PROP. XXIV: THEOR. dies SIMILAR segments of circles upon equal straight See Note. lines, are equal to one another. Let AEB, CFD be similar segments of circles upon the E F B C D . Book-311. because AB is equal to CD: therefore' the straight line AB coinciding with CD, the segment AEB must a coincide with a 23. 3. 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. \ * 10, 1. b 11. 1. c 6.1. с Let ABC be the given segment of a circle; it is required to describe the circle of which it is the segment. Bišecta AC in D, and from the point D drawb DB at right angles to AC, and join AB: first, let the angles ABD, BAD, be equal to one another; then the straight line BD is equal to DA, and therefore to DC; and because the three straight lines DA, DB, DC, are all equal ; D is the centre of the circled: 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 d 9. 3. a : e 23. 1. : is a semicircle: but if the angles ABD, BAD are not equal to one another, at the point A, in the straight line AB, make e the angle BAE equal to the angle ABD, and produce BD, if necessary; to E, and join EC:and because the angle ABE is equal to the angle BAE, the straight line BE is equal to EA: and be cause AD is equal to DC, and DE common to the triangles ADE, CDE, the two sides AD, DE are equal to the two ČD, 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 equals to the base EC: but AE was shown to be equal to EB, wherefore also BE is equal to EC: and the three straight lines 14.1. |