Elements of Geometry |
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Page 26
... semicircle AMDB may be applied exactly to the semicircle ENGF , and then the curved line AMDB will coincide entirely with the curved line ENGF ; but the portion AMD being sup- posed equal to the portion ENG , the point D will fall upon ...
... semicircle AMDB may be applied exactly to the semicircle ENGF , and then the curved line AMDB will coincide entirely with the curved line ENGF ; but the portion AMD being sup- posed equal to the portion ENG , the point D will fall upon ...
Page 35
... semicircle , is a right angle ; for it has for its measure the half of the semicircumference BOD , or the fourth of the circumfer- ence . To demonstrate the same thing in another way , draw the ra dius AC ; the triangle BAC is isosceles ...
... semicircle , is a right angle ; for it has for its measure the half of the semicircumference BOD , or the fourth of the circumfer- ence . To demonstrate the same thing in another way , draw the ra dius AC ; the triangle BAC is isosceles ...
Page 36
... semicir- cumference . And every angle BOC , inscribed in a segment less than a semicircle , is an obtuse angle ; for it has for its measure the half of the arc BAC greater than a semicircumference . 130. Corollary iv . The opposite ...
... semicir- cumference . And every angle BOC , inscribed in a segment less than a semicircle , is an obtuse angle ; for it has for its measure the half of the arc BAC greater than a semicircumference . 130. Corollary iv . The opposite ...
Page 41
... semicircle is a right angle ( 128 ) ; therefore AB , being a perpendicular at the extremity of the radius CB , is a tangent . 152. Scholium . The point A being without the circle , it is evident that there are always two equal tangents ...
... semicircle is a right angle ( 128 ) ; therefore AB , being a perpendicular at the extremity of the radius CB , is a tangent . 152. Scholium . The point A being without the circle , it is evident that there are always two equal tangents ...
Page 42
... to the given angle C. 156. Scholium . If the given angle were a right angle , the segment sought would be a semicircle described upon the diam- eter AB . PROBLEM . 157. To find the numerical ratio of two 42 Elements of Geometry .
... to the given angle C. 156. Scholium . If the given angle were a right angle , the segment sought would be a semicircle described upon the diam- eter AB . PROBLEM . 157. To find the numerical ratio of two 42 Elements of Geometry .
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Common terms and phrases
ABC fig adjacent angles altitude angle ACB angle BAC base ABCD bisect centre chord circ circular sector circumference circumscribed common cone consequently construction convex surface Corollary cube cylinder Demonstration diagonals diameter draw drawn equal angles equiangular equilateral equivalent faces figure formed four right angles frustum GEOM given point gles greater hence homologous sides hypothenuse inclination intersection isosceles triangle JOHN CRERAR LIBRARY join less Let ABC let fall line AC mean proportional measure the half meet multiplied number of sides oblique lines opposite parallelogram parallelopiped perimeter perpendicular plane MN polyedron prism produced proposition radii radius ratio rectangle regular polygon right angles Scholium sector segment semicircle semicircumference side BC similar solid angle sphere spherical polygons spherical triangle square described straight line tangent THEOREM three angles triangle ABC triangular prism triangular pyramids vertex vertices whence