Elements of Geometry |
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Page 41
... inscribed in a semicircle is a right angle ( 128 ) ; therefore AB , being a perpendicular at the extremity of the ... inscribe a circle in a given triangle ABC ( fig . 87 ) . Bisect the angles A and B of the triangle by the straight ...
... inscribed in a semicircle is a right angle ( 128 ) ; therefore AB , being a perpendicular at the extremity of the ... inscribe a circle in a given triangle ABC ( fig . 87 ) . Bisect the angles A and B of the triangle by the straight ...
Page 42
... inscribed in it , shall be equal to a given angle C. Solution . Produce AB toward D , make at the point B the angle ... inscribed angle , has also for its measure the half of the arc AKB ; con- sequently the angle AMB = ABF = EBD = C ...
... inscribed in it , shall be equal to a given angle C. Solution . Produce AB toward D , make at the point B the angle ... inscribed angle , has also for its measure the half of the arc AKB ; con- sequently the angle AMB = ABF = EBD = C ...
Page 67
... inscribed in the same segment ( 127 ) ; for the same reason the angle CB ; therefore these triangles are similar ; and the homologous sides give the proportion AO : DO :: CO : OB . 224. Corollary . Hence AO × OB = DO × CO ; therefore ...
... inscribed in the same segment ( 127 ) ; for the same reason the angle CB ; therefore these triangles are similar ; and the homologous sides give the proportion AO : DO :: CO : OB . 224. Corollary . Hence AO × OB = DO × CO ; therefore ...
Page 69
... inscribed circle . For the triangles AOB , BOC , AOC ( fig . 87 ) , which have Fig . 87 . their common vertex in O , have for their common altitude the radius of the inscribed circle ; consequently the sum of these triangles will be ...
... inscribed circle . For the triangles AOB , BOC , AOC ( fig . 87 ) , which have Fig . 87 . their common vertex in O , have for their common altitude the radius of the inscribed circle ; consequently the sum of these triangles will be ...
Page 70
... inscribed in the same segment AOB ; consequently the triangle ABD is similar to the triangle IBC , and AD : CI :: BD : BC ; whence ADX BC = CI × BD . Again , the triangle ABI is similar to the triangle BDC ; for , the arc AD being equal ...
... inscribed in the same segment AOB ; consequently the triangle ABD is similar to the triangle IBC , and AD : CI :: BD : BC ; whence ADX BC = CI × BD . Again , the triangle ABI is similar to the triangle BDC ; for , the arc AD being equal ...
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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