Elements of Geometry |
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Page 39
... 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 ) . Fig . 87 . Bisect the angles A and B of the triangle by the ...
... 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 ) . Fig . 87 . Bisect the angles A and B of the triangle by the ...
Page 40
... 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 65
... inscribed in the same segment ( 127 ) ; for the same reason the angle C = B ; therefore these triangles are similar , and the homologous sides give the proportion AO : DO :: CO : OB , 224. Corollary . Hence AO x OB = DO × CO ; therefore ...
... inscribed in the same segment ( 127 ) ; for the same reason the angle C = B ; therefore these triangles are similar , and the homologous sides give the proportion AO : DO :: CO : OB , 224. Corollary . Hence AO x OB = DO × CO ; therefore ...
Page 67
... inscribed circle . For the triangles AOB , BOC , AOC ( fig . 87 ) , which have their Fig 87 . common vertex in O , have for their common altitude the radius of the inscribed circle ; consequently the sum of these triangles will be equal ...
... inscribed circle . For the triangles AOB , BOC , AOC ( fig . 87 ) , which have their Fig 87 . common vertex in O , have for their common altitude the radius of the inscribed circle ; consequently the sum of these triangles will be equal ...
Page 68
... inscribed in the same segment AOB ; consequently the triangle ABD is similar to the triangle IBC , and AD : CI :: BD : BC ; whence AD × BC = CI × BD . = Again , the triangle ABI is similar to the triangle BDC ; for the arc AD being ...
... inscribed in the same segment AOB ; consequently the triangle ABD is similar to the triangle IBC , and AD : CI :: BD : BC ; whence AD × BC = CI × BD . = Again , the triangle ABI is similar to the triangle BDC ; for the arc AD being ...
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Common terms and phrases
ABC fig adjacent angles altitude angle ACB angle BAD angles equal 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 and parallel equiangular equilateral equivalent faces figure four right angles frustum Geom gles greater hence homologous sides hypothenuse inclination inscribed circle intersection isosceles join less let fall line AC manner mean proportional measure the half meet multiplied number of sides oblique lines opposite parallelogram parallelopiped perimeter perpendicular plane MN polyedron prism proposition quadrilateral radii radius ratio rectangle regular polygon right angles right-angled triangle Scholium segment semicircumference side BC similar solid angle sphere spherical polygons spherical triangle square described straight line tangent THEOREM three plane angles triangle ABC triangular prism triangular pyramids vertex vertices whence
Popular passages
Page 65 - The square of the hypothenuse is equal to the sum of the squares of the other two sides ; as, 5033 402+302.
Page 21 - If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles are congruent.
Page ii - Co. of the said district, have deposited in this office the title of a book, the right whereof they claim as proprietors, in the words following, to wit : " Tadeuskund, the Last King of the Lenape. An Historical Tale." In conformity to the Act of the Congress of the United States...
Page 63 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. A D A' Hyp. In triangles ABC and A'B'C', To prove AABC A A'B'C' A'B' x A'C ' Proof. Draw the altitudes BD and B'D'.
Page 22 - CIRCLE is a plane figure bounded by a curved line, all the points of which are equally distant from a point within called the centre; as the figure ADB E.
Page ii - States entitled an act for the encouragement of learning hy securing the copies of maps, charts and books to the author., and proprietors of such copies during the times therein mentioned, and also to an act entitled an act supplementary to an act, entitled an act for the encouragement of learning by securing the copies of maps, charts and books to the authors and proprietors of such copies during the times therein mentioned and extending the benefits thereof to the arts of designing, engraving and...
Page 80 - The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides, and their areas are to each other as the squares of those sides (Prop.
Page 164 - If two triangles have two sides and the inchtded angle of the one respectively equal to two sides and the included angle of the other, the two triangles are equal in all respects.
Page 24 - In the same circle, or in equal circles, equal arcs are subtended by equal chords ; and, conversely, equal chords subtend equal arcs.
Page 153 - XVII.) ; hence two similar pyramids are to each other as the cubes of their homologous sides.