 | John Playfair - 1829 - 210 páginas
...manifest that in the same circle, or in equal circles, the jjreater angle stands on the greater arch. CoK. 3. In the same circle, or in equal circles, equal chords are subtended by equal arches. CoR. 4. If a radius ED (Figure 3. 3) bisect a chord AB, it will also bisect... | |
 | Horatio Nelson Robinson - 1860 - 470 páginas
...AE = EB. Hence the theorem. THEOREM III. In the same circle, or in equal circles, equal chords art equally distant from the center. Let AB and EF be equal chords, and C tho. center of the circle. From (7, draw CG and CH, perpendicular to the respective chords. These perpendiculai's... | |
 | Horatio Nelson Robinson - 1865 - 474 páginas
...triangles, ACE, and ECB, are equal in all respects, and AE = EB. Hence the theorem. THEOREM III. In the same circle, or in equal circles, equal chords...These perpendiculars will bisect the chords, (Th. 1), and we shall have AG = EH. "We are now to prove that CG = CH. Since the A's ECU and ACG are right-angled,... | |
 | Horatio Nelson Robinson - 1868 - 276 páginas
...AE — EB. Hence the theorem. THEOREM III. In the same circle, or in equal circles, equal chords art equally distant from the center. Let AB and EF be equal chords, and C tho center of the circle. From (7, draw CGr and CH, perpendicular to the respective chords. These perpendiculars... | |
 | Horatio Nelson Robinson - 1869 - 276 páginas
...triangles, ACE, and ECS, are equal in all respects, and AE = EB. Hence the theorem. THEOREM III. In the same circle, or in equal circles, equal chords...These perpendiculars will bisect the chords, (Th. 1), and we shall have A G- = EH. "We are now to prove that CCr = CH. Since the A's ECH and ACG- are right-angled,... | |
 | Trinity College (Hartford, Conn.) - 1870 - 1008 páginas
...angles on the same side together equal to two right angles, the two lines are parallel. 3. Prove that in the same circle or in equal circles, equal chords are equally distant from the center; and of two unequal chords, the less is the more remote from the center. 4. Prove that if a line is... | |
 | William Chauvenet - 1871 - 380 páginas
...parallel chords is the diameter perpendicular to these chords. PROPOSITION VIII.— THEOREM. 18. 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. 1st. Let AB,... | |
 | William Chauvenet - 1872 - 382 páginas
...parallel chords is the diameter perpendicular to these chords. PROPOSITION VIII.— THEOREM. 18. 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. 1st. Let AB,... | |
 | Association for the improvement of geometrical teaching - 1876 - 66 páginas
...than two straight lines drawn to the circumference are equal, that point is the centre, THEOR. n. In the same circle, or in equal circles, equal chords are equally distant from the centre; and of two, unequal chords the greater is nearer to the centre than the less. [First part by... | |
 | Elias Loomis - 1877 - 458 páginas
...the three sides of a triangle at right angles meet in the same point. PROPOSITION VIII. THEOREM.. In the same circle or in equal circles, equal chords are equally distant from the centre; and of two unequal chords, the less is the more remote from the centre. Let the chords AB,... | |
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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. 
