Elements of Geometry |
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Page 73
... equal . Hence , all its diameters are equal , since the diameter is equal to twice the radius . Fig . 1 . M M A B A A B P Fig . 2 . D Fig . 3 . 164. DEF . An Arc of a circle is any portion of the circum- ference , as A MB , Fig . 3 ...
... equal . Hence , all its diameters are equal , since the diameter is equal to twice the radius . Fig . 1 . M M A B A A B P Fig . 2 . D Fig . 3 . 164. DEF . An Arc of a circle is any portion of the circum- ference , as A MB , Fig . 3 ...
Page 78
... equal circles , equal arcs subtend equal angles at the centre . R S Α ' A B A R BI P P In the equal circles ABP and A'B ' P ' let arc RS = arc R ' S ' . We are to prove ROSZ R ' O'S ' . Apply O ABP to O A'B ' P ' , so that the radius ...
... equal circles , equal arcs subtend equal angles at the centre . R S Α ' A B A R BI P P In the equal circles ABP and A'B ' P ' let arc RS = arc R ' S ' . We are to prove ROSZ R ' O'S ' . Apply O ABP to O A'B ' P ' , so that the radius ...
Page 80
... equal circles , equal chords subtend equal arcs . A R S B A R BI P PI In the equal circles ABP and A'B ' P ' , let ... radius perpendicular to 80 GEOMETRY . BOOK II .
... equal circles , equal chords subtend equal arcs . A R S B A R BI P PI In the equal circles ABP and A'B ' P ' , let ... radius perpendicular to 80 GEOMETRY . BOOK II .
Page 81
... radius CS be per- pendicular to AB at the point M. We are to prove A M = BM , and arc AS = arc B S. Draw CA and C B. СА = СВ , ( being radii of the same ) ; ..A ACB is isosceles , .. ( the opposite sides being equal ) ; CS bisects the ...
... radius CS be per- pendicular to AB at the point M. We are to prove A M = BM , and arc AS = arc B S. Draw CA and C B. СА = СВ , ( being radii of the same ) ; ..A ACB is isosceles , .. ( the opposite sides being equal ) ; CS bisects the ...
Page 103
... radius equal to n , describe an arc . From B as a centre , with a radius equal to o , describe an arc intersecting the former arc at C. C is the required point . Q. E. F. 216. COROLLARY 1. By continuing these arcs , another point below ...
... radius equal to n , describe an arc . From B as a centre , with a radius equal to o , describe an arc intersecting the former arc at C. C is the required point . Q. E. F. 216. COROLLARY 1. By continuing these arcs , another point below ...
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Common terms and phrases
A B C D AABC AACB AB² ABCD adjacent angles apothem arc A B base and altitude BC² centre centre of symmetry circumference circumscribed construct a square COROLLARY decagon diagonals diameter divided Draw equal arcs equal distances equal respectively equiangular equiangular polygon equilateral equilateral polygon exterior angles figure given circle given line given polygon given square homologous sides hypotenuse intersecting isosceles Let A B Let ABC line A B measured by arc middle point number of sides parallelogram perpendicular plane polygon ABC polygon similar PROBLEM prove Q. E. D. PROPOSITION quadrilateral radii radius equal ratio rect rectangles regular inscribed regular polygon required to construct right angles right triangle SCHOLIUM segment semicircle similar polygons subtend symmetrical with respect tangent THEOREM triangle ABC vertex vertices
Popular passages
Page 40 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz.
Page 126 - To describe an isosceles triangle having each of the angles at the base double of the third angle.
Page 136 - The first of four magnitudes is said to have the same ratio to the second which the third has to the fourth, when...
Page 207 - Construct a rectangle having the difference of its base and altitude equal to a given line, and its area equivalent to the sum of a given triangle and a given pentagon.
Page 202 - In any proportion, the product of the means is equal to the product of the extremes.
Page 142 - If a line divides two sides of a triangle proportionally, it is parallel to the third side.
Page 175 - Any two rectangles are to each other as the products of their bases by their altitudes.
Page 72 - Every point in the bisector of an angle is equally distant from the sides of the angle ; and every point not in the bisector, but within the angle, is unequally distant from the sides of the angle.
Page 73 - A 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 146 - 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'.