Elements of Geometry |
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Page vi
... construction of figures . The third section , entitled the proportions of figures , contains the measure of surfaces , their comparison , the properties of a right - angled triangle , those of equiangular triangles , of similar figures ...
... construction of figures . The third section , entitled the proportions of figures , contains the measure of surfaces , their comparison , the properties of a right - angled triangle , those of equiangular triangles , of similar figures ...
Page 9
... construction ; therefore ( 43 ) the angle B is equal to the angle C. 46. Corollary . An equilateral triangle is also equiangular , that is , it has its angles equal . Geom . 2 47. Scholium . From the equality of the triangles ABD Of ...
... construction ; therefore ( 43 ) the angle B is equal to the angle C. 46. Corollary . An equilateral triangle is also equiangular , that is , it has its angles equal . Geom . 2 47. Scholium . From the equality of the triangles ABD Of ...
Page 15
... construction , a right angle , conse quently the straight lines AC , BD , are perpendicular to the same straight line EF , therefore they are parallel ( 57 ) . THEOREM . 61. If two straight lines , AI , BD ( fig . 36 ) , make with a ...
... construction , a right angle , conse quently the straight lines AC , BD , are perpendicular to the same straight line EF , therefore they are parallel ( 57 ) . THEOREM . 61. If two straight lines , AI , BD ( fig . 36 ) , make with a ...
Page 35
... construction will serve to make a right angle BAD at a given point A in a given line BC . PROBLEM . 135. From a given point A ( fig . 72 ) without the straight line BD , Fig . 72 . to let fall a perpendicular upon this line , Solution ...
... construction will serve to make a right angle BAD at a given point A in a given line BC . PROBLEM . 135. From a given point A ( fig . 72 ) without the straight line BD , Fig . 72 . to let fall a perpendicular upon this line , Solution ...
Page 36
... construction , we may bisect each of the halves AE , EB , and thus , by successive subdivis- ions , we may divide an angle or arc into four , eight , sixteen , & c . , equal parts . PROBLEM . 139. Through a given point A ( fig . 75 ) ...
... construction , we may bisect each of the halves AE , EB , and thus , by successive subdivis- ions , we may divide an angle or arc into four , eight , sixteen , & c . , equal parts . PROBLEM . 139. Through a given point A ( fig . 75 ) ...
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Common terms and phrases
ABC fig adjacent angles altitude angle ACB angle BAD 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 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 pyramid S-ABC 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