Elementary Synthetic GeometryJ. Wiley & sons, 1896 - 164 Seiten |
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... Chords , VII . Two Circles , SECTIONS • I - 54 55-109 110-119 • · • 120-152 153-171 172-183 184-186 • 187-205 206-227 228-243 • 244-261 262-268 · • 269-280 • • 281-299 • · • 300-313 VIII . Parallels , • IX . The Triangle , X. Polygons ...
... Chords , VII . Two Circles , SECTIONS • I - 54 55-109 110-119 • · • 120-152 153-171 172-183 184-186 • 187-205 206-227 228-243 • 244-261 262-268 · • 269-280 • • 281-299 • · • 300-313 VIII . Parallels , • IX . The Triangle , X. Polygons ...
Seite 10
... chord . on the 76. Any chord through the center is called a diameter . 77. All diameters are equal , each being equal to two radii . 78. Every diameter is bisected by the center of the circle . 79. No circle can have more than one ...
... chord . on the 76. Any chord through the center is called a diameter . 77. All diameters are equal , each being equal to two radii . 78. Every diameter is bisected by the center of the circle . 79. No circle can have more than one ...
Seite 14
... chord . FIG . 44 . A Proof . For by folding over along CO we bring A into coin- cidence with A ' . Therefore sect AM = sect A'M . OMA = A'MO . ¥ MCA = ¥ A'CM . AOM = × MOA ' . T CHAPTER III . THE FUNDAMENTAL PROBLEMS . 110. Problem 14 ...
... chord . FIG . 44 . A Proof . For by folding over along CO we bring A into coin- cidence with A ' . Therefore sect AM = sect A'M . OMA = A'MO . ¥ MCA = ¥ A'CM . AOM = × MOA ' . T CHAPTER III . THE FUNDAMENTAL PROBLEMS . 110. Problem 14 ...
Seite 15
... chord , and the center of the circle , is per- pendicular to the chord , and bisects the exple- mental arcs , and their angles at the center . Proof . A and B are any points on 0 . Turn the whole figure over through half a revo- lution ...
... chord , and the center of the circle , is per- pendicular to the chord , and bisects the exple- mental arcs , and their angles at the center . Proof . A and B are any points on 0 . Turn the whole figure over through half a revo- lution ...
Seite 16
... chord AA ' , and join its mid point , M , to the center C. ACA ' . 113. Problem . At a given point on a given straight , to draw a perpendicular to that straight . Construction . Bisect the straight angle at the point . 114. Problem ...
... chord AA ' , and join its mid point , M , to the center C. ACA ' . 113. Problem . At a given point on a given straight , to draw a perpendicular to that straight . Construction . Bisect the straight angle at the point . 114. Problem ...
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Häufige Begriffe und Wortgruppen
altitude angle equal angle-bisector angles opposite axis of symmetry bisects Brocard p't called cenquad circumcenter co-straight coincide common point concurrent concyclic congruent Construction Corollary corresponding diagonal diameter draw end point Engineering equal angles equal sects equal sides equiangular polygon explemental exterior angle figure g-line Geometry given p't given point given straight greater harmonic conjugates harmonic range inscribed inscribed angles intercept intersection point inversely isogonal isogonal conjugates Lemoine p't Lemoine point lune magnitude median mid p't mid point morocco opposite sides orthocenter parallel parallelogram pencil perigon perspective plane polar pole polygon Proof quad quadrant quadrilateral radius rectangle rhombus right angles rotation sect joining semicircle sides equal sphere spherical angle spherical polygon spherical triangle square st's straight angle supplemental symcenter symcentral symmedians symmetrical with regard symtra tangent Theorem third side transversal vertex vertices
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