 The perpendicular bisector of a chord passes through the center of the circle and bisects the arcs subtended by the chord.  New Plane Geometry - Seite 121von Wooster Woodruff Beman, David Eugene Smith - 1899 - 252 SeitenVollansicht - Über dieses Buch
 | Wooster Woodruff Beman, David Eugene Smith - 1895 - 344 Seiten
...perpendicular to it. ForvAE=EC, and OA = OC, and OE = OE, .-. A AEO-S A CEO, and Z OEA = Z CEO, by I, th. 12. 2. The perpendicular bisector of a chord passes through...center of the circle and bisects the subtended arcs. Theorem 6. All points in a chord lie within the circle; and all points in the same line, but not in... | |
 | Wooster Woodruff Beman, David Eugene Smith - 1895 - 346 Seiten
...to it. For v AE = EC, and OA = OC, and OE = OE, .'. A AEOSS A CEO, and Z OEA = Z CEO, by I, th. 12. 2. The perpendicular bisector of a chord passes through...center of the circle and bisects the subtended arcs. For the center is equidistant from the ends of the chord, by definition of a circle ; .'. it lies on... | |
 | Wooster Woodruff Beman, David Eugene Smith - 1899 - 412 Seiten
...OC, .'. DB has two points equidistant from A and C. Hence, being determined by these points, it is .L to AC, by I, prop. XLI. 2. The perpendicular bisector...center of the circle and bisects the subtended arcs. For the center is equidistant from the ends of the chord, by definition of a circle ; .'. it lies on... | |
 | Wooster Woodruff Beman, David Eugene Smith - 1900 - 395 Seiten
...has two points equidistant from A and C. Hence, being determined by these points, it is J- to A (7, by I, prop. XLI. 2. The perpendicular bisector of...line, but not in the chord, lie without the circle. Given the points Pl in a chord AB, and P2 in AB produced. To prove that P1 is within the circle, and... | |
 | Arthur Schultze - 1901 - 392 Seiten
...is the usual means of proving : (1) The equality of lines ? (2) The equality of arcs ? 179. COR. A perpendicular bisector of a chord passes through the center of the circle. Ex. 313. If two chords are equal, the perpendiculars from the center upon the chords are equal. Ex.... | |
 | Arthur Schultze, Frank Louis Sevenoak - 1901 - 396 Seiten
...is the usual means of proving : (1) The equality of lines ? (2) The equality of arcs ? 179. COR. A perpendicular bisector of a chord passes through the center of the circle. Ex. 313. If two chords are equal, the perpendiculars from the center upon the chords are equal. Ex.... | |
 | Arthur Schultze - 1901 - 260 Seiten
...is the usual means of proving : (1) The equality of lines ? (2) The equality of arcs ? 179. COR. A perpendicular bisector of a chord passes through the center of the circle. Ex. 313. If two chords are equal, the perpendiculars from the center upon the chords are equal. Ex.... | |
 | Arthur Schultze, Frank Louis Sevenoak - 1902 - 394 Seiten
...is the usual means of proving : (1) The equality of lines ? (2) The equality of arcs ? 179. COR. A perpendicular bisector of a chord passes 'through the center of the circle. Ex. 313. If two chords are equal, the perpendiculars from the center upon the chords are equal. Ex.... | |
 | Wooster Woodruff Beman, David Eugene Smith - 1903 - 400 Seiten
...OC. .'. DB has two points equidistant from A and C. Hence, being determined by these points, it is J_ to AC, by I, prop. XLI. 2. The perpendicular bisector...the center of the circle and bisects the subtended ares. PROPOSITION VI. 185. Theorem. All points in a chord lie within the circle ; and all points in... | |
 | Fletcher Durell - 1911 - 553 Seiten
...A diameter which bisects a chord (shorter than a diameter) is perpendicular to the chord. 223. COR. 2. The perpendicular bisector of a chord passes through the center of the circle, and bisects the arcs subtended by the chord. 224. COR. 3. A line from the center perpendicular to a chord bisects the... | |
| |
