PROPOSITION II. THEOREM. 228. A straight line cannot intersect the circumference of a circle in more than two points. Let HK be any line cutting the circumference AMP. To prove that HK can intersect the circumference in only two points. Proof. If possible, let HK intersect the circumference in three points H, P, and K. From O, the centre of the O, draw OH, OP, and OK. Then OH, OP, and OK are equal, (being radii of the same circle). Hence, we have three equal straight lines OH, OP, and OK drawn from the same point to a given straight line. But this is impossible, § 120 (only two equal straight lines can be drawn from a point to a straight line). Therefore, HK can intersect the circumference in only two points. Q. E. D. PROPOSITION III. THEOREM. 229. In the same circle, or equal circles, equal angles at the centre intercept equal arcs; CONVERSELY, equal arcs subtend equal angles at the centre. In the equal circles ABP and A'B'P' let 20 = 20. (for OR O'R', and OS = O'S', being radii of equal ©). Then the arc RS will coincide with the arc R'S', since all points in the arcs are equidistant from the centre. Proof. Apply O ABP to O A'B'P', so that are RS shall fall upon arc R'S', R falling upon R', S upon S', and O upon O'. Then RO will coincide with R'O', and SO with S'O'. .. O and O' coincide and are equal. Q. E. D. PROPOSITION IV. THEOREM. 230. In the same circle, or equal circles, equal chords subtend equal arcs; CONVERSELY, equal arcs are subtended by equal chords. In the equal circles ABP and A'B'P', let chord_RS = chord R'S'. the radii OR and OS the radii O'R' and O'S'. § 226 = ..A ROSA R'O'S', (three sides of the one being equal to three sides of the other). (in equal, equal at the centre intercept equal arcs). Let arc RS: = arc R'S'. chord RS chord R'S'. = § 160 $ 229 Q. E. D. CONVERSELY: To prove LO', § 229 (equal arcs in equal © subtend equal ▲ at the centre), and OR and OS O'R' and O'S', respectively. § 226 ..▲ ORS=▲ O'R'S', $150 (having two sides equal each to each and the included & equal). .. chord RS= chord R'S'. Q. E. D. PROPOSITION V. THEOREM. 231. In the same circle, or equal circles, if two arcs are unequal, and each is less than a semi-circumference, the greater arc is subtended by the greater chord; CONVERSELY, the greater chord subtends the greater arc. A M In the circle whose centre is 0, let the arc AMB be greater than the arc AMF. To prove Proof. chord AB greater than chord AF. Draw the radii OA, OF, and OB. Since Fis between A and B, OF will fall between OA and OB, and ZAOB be greater than ▲ AOF. Hence, in the AAOB and AOF, the radii OA and OB = the radii OA and OF, but AOB is greater than ▲ AOF. .. AB > AF, $ 152 (the having two sides equal each to each, but the included ▲ unequal). CONVERSELY: Let AB be greater than AF. To prove arc AB greater than arc AF. In the AAOB and AOF OA and OB = OA and OF respectively. But AB is greater than AF. :. ZAOB is greater than ▲ AOF, Hyp. § 153 (the having two sides equal each to each, but the third sides unequal). .. OB falls without OF ..arc AB is greater than arc AF. Q. E. D. PROPOSITION VI. THEOREM. 232. The radius perpendicular to a chord bisects the chord and the arc subtended by it. E M Let AB be the chord, and let the radius OS be perpendicular to AB at M. To prove AM= BM, and arc AS= arc BS. Proof. Draw OA and OB from O, the centre of the circle. In the rt. A OAM and OBM the radius OA = the radius OB, and OM OM. ..ΔΟΑΜ=ΔΟΒΜ, Iden. § 161 (having the hypotenuse and a side of one equal to the hypotenuse and a and side of the other). :. AM=BM, AOS= L BOS. .. arc AS= arc BS, (equal at the centre intercept equal arcs on the circumference). Q. E. D. 233. COR. 1. The perpendicular erected at the middle of a chord passes through the centre. For the centre is equidistant from the extremities of a chord, and is therefore in the perpendicular erected at the middle of the chord. $122 234. COR. 2. The perpendicular erected at the middle of a chord bisects the arc of the chord. 235. COR. 3. The locus of the middle points of a system of parallel chords is the diameter perpendicular to them. |