 | George Albert Wentworth - Geometry - 1899 - 496 pages
...bisects a chord is perpendicular to it. 82 ARCS, CHORDS, AND TANGENTS. PROPOSITION VI. THEOREM. 249. In the same circle or in equal circles, equal chords are equally distant from the centre. CONVERSELY : Chords equally distant from the centre are equal. Let AB and CF be equal chords... | |
 | Webster Wells - Geometry - 1899 - 450 pages
...subtended arc is perpendicular to the chord. (By § 158, chord AC = chord BC.) PROP. X. THEOREM. 164. In the same circle, or in equal circles, equal chords are equally distant from the centre. Given AB and CD equal chords of O ABC, whose centre is O, and lines OE and OF _L to AB and... | |
 | George Albert Wentworth - Geometry, Solid - 1899 - 248 pages
...angle. 245. A diameter perpendicular to a chord bisects the chord and the arcs subtended by it. 249. In the same circle or in equal circles, equal chords are equally distant from the centre. CONVERSELY : Chords equally distant from the centre are equal. 253. A straight line perpendicular... | |
 | Harvard University - Geometry - 1899 - 39 pages
...the straight line joining their centres bisects at right angles their common chord. THEOREM IX. In the same circle or in equal circles, equal chords are equally distant from the centre ; and of two unequal chords the less is at the greater distance from the centre. Conversely,... | |
 | Arkansas. State Department of Public Instruction - Education - 1900 - 236 pages
...nonparallel sides of a trapezoid is parallel to the bases and is equal to one-half their sum. 7. In the same circle, or in equal circles, equal chords are equally distant from the center; conversely, chords equally distant from the center are equal. 8. To describe upon a given straight... | |
 | Arthur Schultze - 1901 - 260 pages
...184. COR. 4. Two circumferences cannot meet in more thsji two points. PROPOSITION V THEOREM 185. In the same circle, or in equal circles, equal chords are equally distant from the center; and, conversely, chords equally distant from the center are equal. B Hyp. —InO^-BCD: D chord AB =... | |
 | Arthur Schultze, Frank Louis Sevenoak - Geometry - 1901 - 396 pages
...center of a given circle. Ex. 318. To find the midpoint of a given arc. PROPOSITION V THEOREM 185. In the same circle, or in equal circles, equal chords are equally distant from the center; and, conversely, chords equally distant from the center are equal. D Hyp. — InOABCZ): chord AB =... | |
 | Arthur Schultze - 1901 - 392 pages
...184. COR. 4. Two circumferences cannot meet in more than two points. PROPOSITION V THEOREM 185. In the same circle, or in equal circles, equal chords are equally distant from the center ; and, conversely, chords equally distant from the center are equal. Hyp. — InOABC£>: chord AB =... | |
 | Alan Sanders - Geometry, Modern - 1901 - 260 pages
...of these arcs, prove that it is perpendicular to AB and bisects it. PROPOSITION X. THEOREM 283. In the same circle or in equal circles equal chords are equally distant from the center; and conversely, chords that are equally distant from the center are equal. Let AB and CD be equal chords... | |
 | Edward Brooks - Geometry, Modern - 1901 - 278 pages
...There can be but one point equally distant from three other points. PROPOSITION X. — THEOREM. In the same circle, or in equal circles, equal chords are equally distant from the centre, and CONVERSELY. Given.— In the circle ADB let the chords AB and CD be equal. To Prove. —... | |
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In the same circle, or in equal circles, equal chords are equally distant from the center; and, conversely, chords equally distant from the center are equal. 
