 | Edward Rutledge Robbins - Geometry, Plane - 1906 - 268 pages
...point B on M (?) (201). Arc AB will coincide with arc LM (?) (192). .-. AB = LM (?) (28). QED 207. THEOREM. In the same circle (or in equal circles) equal arcs are intercepted by equal central angles. [Converse.] Given : OO = OC;arc.4B = arc LM. To Prove : Z o =... | |
 | Edward Rutledge Robbins - Geometry - 1907 - 428 pages
...and point B on M (?) (201). Arc AB will coincide with arc LM (?) (192). ...AB = LM (?) (28). QED 207. THEOREM. In the same circle (or in equal circles) equal arcs are intercepted by equal central angles. [Converse.] Given : OO = OC ; arc AB = arc LM. To Prove : ZO =... | |
 | George William Myers - Mathematics - 1910 - 304 pages
...form unequal angles intercept unequal arcs on the circumference, and conversely. PROPOSITION XXVI 95. Theorem: In the same circle or in equal circles, equal...arcs are subtended by equal chords, and conversely. I. Given circle O, with arc AB=arc CD (Fig. 63). FIG. 63 To prove chord AB =chord C D. This is left... | |
 | William Herschel Bruce, Claude Carr Cody (Jr.) - Geometry, Modern - 1910 - 284 pages
...or in equal circles, sectors with equal central angles are equal. PROPOSITION II. THEOBEM. 233. In the same circle, or in equal circles, equal arcs are subtended ~by equal chords and intercepted by equal central angles. Given OC = O C' and arc AB = arc A'B'. To prove chord AB= chord... | |
 | Herbert Ellsworth Slaught, Nels Johann Lennes - Geometry, Plane - 1910 - 300 pages
...and Z. C coincides with Z C'. Then A falls on A1 and B on B'. (§ 189) Hence AB = A!B'. (§ 196) 200. THEOREM. In the same circle or in equal circles equal arcs are intercepted by equal central angks. Given OC = OC, AB = A'ff. (See figure, § 199.) To prove that Z.... | |
 | Herbert Ellsworth Slaught, Nels Johann Lennes - Geometry, Plane - 1910 - 304 pages
...Cf, and ZC coincides with Z c'. Then A falls on A' and B on B'. (§ 189) Hence AB = AW. (§ 196) 200. THEOREM. In the same circle or in equal circles equal arcs are intercepted by equal central angles. Given QC=QC,AB = A'ff. (See figure, § 199.) To prove that Z c=... | |
 | Arthur Schultze, Frank Louis Sevenoak - Geometry, Modern - 1911 - 266 pages
...circumference. The circle is then said to be circumscribed about the polygon. PROPOSITION II. THEOKEM 177. In the same circle, or in equal circles, equal arcs are...and, conversely, equal chords subtend equal arcs. Hyp. In equal ©, 0 and 0', Proof. Draw radii, OA, OB, O'A', O'B'. Z AOB = Z A'O'B', (equal arcs subtend... | |
 | Robert Louis Short, William Harris Elson - Mathematics - 1911 - 216 pages
...to be inscribed in the segment ABC. The angle A is inscribed in segment CAB. THEOREM LVIII 244. In the same circle or in equal circles, equal arcs are subtended by equal chords. Use figure in § 238. Draw chords BA and B'A'. Prove A OAB = A O'A'B'. 245. The converse of this theorem... | |
 | Education - 1904 - 484 pages
...This and similar exercises may be assigned as home work. EXERCISE 79. Show by superposition that in the same circle, or in equal circles, equal arcs are subtended by equal chords; that equal chords are equidistant from the centre. After following the directions of the above in several... | |
 | William Herschel Bruce, Claude Carr Cody - Geometry, Solid - 1912 - 134 pages
...the included angle of the other. 213. Radii of the same circle or of equal circles are equal. 233. In the same circle, or in equal circles, equal arcs are subtended by equal chords and intercepted by equal central angles. 234. In the same circle, or in equal circles, equal chords subtend... | |
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THEOREM. In the same circle, or in equal circles, equal arcs are subtended by equal chords ; and, conversely, equal chords subtend equal arcs. 
