Chauvenet's Treatise on Elementary Geometry |
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Page 54
... arc equal to half a circum- ference , as the arc AMB , is called a semi - circumference . PROPOSITION III . - THEOREM . 9. A diameter is greater than any other chord . Let AC be any chord which is not a diame- ter , and AOB a diameter ...
... arc equal to half a circum- ference , as the arc AMB , is called a semi - circumference . PROPOSITION III . - THEOREM . 9. A diameter is greater than any other chord . Let AC be any chord which is not a diame- ter , and AOB a diameter ...
Page 55
... . 13. In equal circles , or in the same circle , the greater arc is subtended by the greater chord , and conversely ; the arcs being both less than a semi - circumference . Let the arc AC be greater than the arc AB BOOK II . 55.
... . 13. In equal circles , or in the same circle , the greater arc is subtended by the greater chord , and conversely ; the arcs being both less than a semi - circumference . Let the arc AC be greater than the arc AB BOOK II . 55.
Page 56
William Chauvenet. Let the arc AC be greater than the arc AB ; then , the chord AC is greater than the chord AB . M M B C For , draw the radii OA , OB , OC . In the triangles AOC , A OB , the angle AOC is obviously greater than the ...
William Chauvenet. Let the arc AC be greater than the arc AB ; then , the chord AC is greater than the chord AB . M M B C For , draw the radii OA , OB , OC . In the triangles AOC , A OB , the angle AOC is obviously greater than the ...
Page 61
... AC and BD are equal . For , let OM be the radius drawn perpendicular to the parallels . By Prop . VII . the point M is at once the middle of the arc AMB and of the arc CMD , and hence we have G K N H AM BM and CM = DM , whence , by ...
... AC and BD are equal . For , let OM be the radius drawn perpendicular to the parallels . By Prop . VII . the point M is at once the middle of the arc AMB and of the arc CMD , and hence we have G K N H AM BM and CM = DM , whence , by ...
Page 60
... AC and BD are equal . For , let OM be the radius drawn perpendicular to the parallels . By Prop . VII . the point M is at once the middle of the arc AMB and of the arc CMD , and hence we have G N H AM = BM and CM = DM , whence , by ...
... AC and BD are equal . For , let OM be the radius drawn perpendicular to the parallels . By Prop . VII . the point M is at once the middle of the arc AMB and of the arc CMD , and hence we have G N H AM = BM and CM = DM , whence , by ...
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Common terms and phrases
ABCD adjacent angles allel angle AOB angle BAC angles are equal apothem arc AC BC² bisector bisects called centre chord circum circumference coincide Corollary Definition diagonals diameter dicular distance divided draw equal angles equally distant exterior angle ference figure four right angles geometry given angle given point given straight line greater hence homologous sides hypotenuse inscribed inscribed angle intercepted arcs less Let ABC measured by one-half middle point number of sides oblique lines one-half the arc opposite sides parallel to BC parallelogram pendicular perimeter perpen perpendicular points of intersection produced proportion PROPOSITION quadrilateral quantities radii radius ratio rectangle regular polygon rhombus right angles right triangles Scholium secant line segment semi-circumference side AC sides are equal square straight line drawn symmetrical with respect tangent theorem triangle ABC triangles are equal vertex
Popular passages
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