| Benjamin Peirce - Geometry - 1837 - 216 pages
...AB. The triangle ABC is therefore isosceles, by art. 58, and the chords AB and BC are equal. ' 112. Theorem. Conversely, in the same circle, or in equal circles, equal chords subtend equal arcs. Perpendicular at the Middle of a Chord. Demonstration. Let the chord AB (fig. 52) be equal to thejifcord... | |
| Benjamin Peirce - Geometry - 1841 - 186 pages
...for its measure the half of an arc BEC less than a semicircumference. 111. Corollary. Every angle BEC inscribed in a segment less than a semicircle is an...A and C must be equal, by § 55, and also the arcs ABt and BC, which are double their measures. 114. Theorem. In the same circle, or in equal circles,... | |
| Benjamin Peirce - Geometry - 1855 - 184 pages
...the arc JHt (fig. 52) be equal to the arc BC. Join AC ; and, in the triangle ABC, the angles A and 0 are equal, for they are measured by the halves of...Proof. Let the chord AB (fig. 52) be equal to the chord J5C. Join AC ; and in the isosceles triangle ABC the angles A and C must be equal, by § 55, and also... | |
| Benjamin Peirce - 1868 - 200 pages
...triangle ABC, the angles A and C are equal, for they are measured by the halves of the equal arcs JBC and AB. The triangle ABC is therefore isosceles, by...chord BC. Join AC ; and in the isosceles triangle .ABCthe angles A and C must be equal, by § 55, and also the arcs AB and BC, which are double their... | |
| Benjamin Peirce - 1870 - 200 pages
...for its measure the half of an arc BEC less than a semicircumference. 111. Corollary. Every angle BEC inscribed in a segment less than a semicircle is an...Proof. Let the chord AB (fig. 52) be equal to the chord JBC. Join AC ; and in the isosceles triangle .ABC the angles A and C must be equal, by § 55, and also... | |
| Benjamin Peirce - Geometry - 1873 - 202 pages
...of the equal arcs BC and AB. The triangle JIBC is therefore isosceles, by ^ 58, and the chords JLB and BC are equal. 113. Theorem. Conversely, in the...the arcs AB and BC, which are double their measures. 1 14. Theorem. In the same circle, or in equal circles, if the sum of two* arcs be less than a circumference,... | |
| George Anthony Hill - Geometry - 1889 - 200 pages
...is said "to be subtended by" the chord AB. To subtend means to stretch across. 18. Theorem. — In the same circle, or in equal circles, equal chords subtend equal arcs. What is the hypothesis? What is the conclusion ? PROOF. Apply No. 1, p. 70, and No. 15, p. 96. Fio.... | |
| George Anthony Hill - 1891 - 206 pages
...is said "to be subtended by" the chord AB. To subtend means to stretch across. 18. Theorem. — In the same circle, or in equal circles, equal chords subtend equal arcs. "What is the hypothesis? What is the conclusion ? PROOF. Apply No. 1, p. 70, and No. 15, p. 96. Fio.... | |
| George D. Pettee - Geometry, Plane - 1896 - 272 pages
...of = 'S 1 ] coincident extremities [common vertex, coincident sides] PROPOSITION II 146. Theorem. In the same circle, or in equal circles, equal chords subtend equal arcs, and, conversely, equal arcs are subtended by equal chords. Appl. Let m represent the n " " AB=12. Prove... | |
| George Washington Hull - Geometry - 1897 - 408 pages
...concave arc is the line which passes through the centre of the circle. PROPOSITION VI. THEOREM. 155. In the same circle, or in equal circles, equal chords subtend equal arcs. D Given—AC and MN equal chords in the equal circles A CD and MNP. To Prove— arc AC= arc MN. Dem.—Draw... | |
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