Elementary Geometry, Τόμος 2 |
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Σελίδα 4
... Hence it follows logically that in a circle , ( 4 ) , angles at the centre , ( B ) , which are equal ( D ) , stand on equal arcs , ( C ) . All converse theorems can be at once proved by the reductio ad absurdum , but we shall in general ...
... Hence it follows logically that in a circle , ( 4 ) , angles at the centre , ( B ) , which are equal ( D ) , stand on equal arcs , ( C ) . All converse theorems can be at once proved by the reductio ad absurdum , but we shall in general ...
Σελίδα 5
... Hence the semicircle APB if folded over the line AB , would coincide with the semicircle AQB , PN falling on NQ ; and therefore the arc AP = the arc AQ , and the arc PB the arc BQ . = COR . 2. Hence also parallel chords in a circle ...
... Hence the semicircle APB if folded over the line AB , would coincide with the semicircle AQB , PN falling on NQ ; and therefore the arc AP = the arc AQ , and the arc PB the arc BQ . = COR . 2. Hence also parallel chords in a circle ...
Σελίδα 6
James Maurice Wilson. COR . 2. Hence also parallel chords in a circle intercept equal arcs . = If PQ , RS are parallel ... Hence ( B ) If a line bisects a chord at right angles ( 1 ) and ( 3 ) , it must pass through the centre ( 2 ) . And ...
James Maurice Wilson. COR . 2. Hence also parallel chords in a circle intercept equal arcs . = If PQ , RS are parallel ... Hence ( B ) If a line bisects a chord at right angles ( 1 ) and ( 3 ) , it must pass through the centre ( 2 ) . And ...
Σελίδα 8
... Hence the centre must be at O , the point of intersection of these perpendiculars ; and the circle described with centre O and radius OA will pass through A , B and C. And there can be only one centre , since the perpen- diculars ...
... Hence the centre must be at O , the point of intersection of these perpendiculars ; and the circle described with centre O and radius OA will pass through A , B and C. And there can be only one centre , since the perpen- diculars ...
Σελίδα 9
... Hence , ( 1 ) if AB = CD and AN = CM , it follows that OM = ON . ( 2 ) If AB > CD , AN > CM , and therefore OM is ON . ( 3 ) If ON = OM , therefore AN = CM and AB = CD . ( 4 ) If ON OM , AN > CM and AB > CD . COR . I. The diameter is ...
... Hence , ( 1 ) if AB = CD and AN = CM , it follows that OM = ON . ( 2 ) If AB > CD , AN > CM , and therefore OM is ON . ( 3 ) If ON = OM , therefore AN = CM and AB = CD . ( 4 ) If ON OM , AN > CM and AB > CD . COR . I. The diameter is ...
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Συχνά εμφανιζόμενοι όροι και φράσεις
ABCD Algebra angle AOB angular points ARITHMETIC base bisectors bisects the angle BROOKE FOSS WESTCOTT Cambridge centre circles touch circumference circumscribed cloth common measure common tangents Construction Crown 8vo decagon Describe a circle diameter divided drawn EDUCATIONAL BOOKS ELEMENTARY TREATISE equal arcs equal chords equiangular equilateral triangle Examples Fcap find the locus follows GEOMETRY given circle given line given point given ratio given straight line Hence homologous hypothenuse incommensurable inscribed intersect John's College Join Let ABC LIST OF EDUCATIONAL magnitudes Mathematical mean proportional middle points numerical values parallel parallelograms pass perimeter perpendicular point of contact PROBLEM prove Ptolemy's Theorem quadrilateral figure radii radius rectangle contained regular polygon right angles Schools secant Second Edition segments shew shewn similar triangles SPQR student subtend equal angles tangent THEOREM triangle ABC vertical angle
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