Elements of Geometry |
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Page 73
... subtends two arcs whose sum is the cir- cumference . Thus the chord AB , ( Fig . 3 ) , subtends the arc AMB and the arc A D B. Whenever a chord and its arc are spoken of , the less arc is meant unless it be otherwise stated . 167. DEF ...
... subtends two arcs whose sum is the cir- cumference . Thus the chord AB , ( Fig . 3 ) , subtends the arc AMB and the arc A D B. Whenever a chord and its arc are spoken of , the less arc is meant unless it be otherwise stated . 167. DEF ...
Page 78
... 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 O R shall fall upon O ' R ...
... 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 O R shall fall upon O ' R ...
Page 79
... subtended by equal chords . 4 པ R S R B A BI P P In the equal circles ABP and A'B ' P ' let arc RS = arc R ' S ' . We are to ... subtend equal at the centre ) . § 176 $ 176 $ 180 § 106 ( two sides and the included of the one being equal ...
... subtended by equal chords . 4 པ R S R B A BI P P In the equal circles ABP and A'B ' P ' let arc RS = arc R ' S ' . We are to ... subtend equal at the centre ) . § 176 $ 176 $ 180 § 106 ( two sides and the included of the one being equal ...
Page 80
... subtend equal arcs . A R S B A R BI P PI In the equal circles ABP and A'B ' P ' , let chord RS = chord R'S ' . We are to prove = arc RS arc R ' S ' . Draw the radii O R , O S , O ' R ' , and O ' S ' . In the ROS and R ' O'S ' RS = R'S ...
... subtend equal arcs . A R S B A R BI P PI In the equal circles ABP and A'B ' P ' , let chord RS = chord R'S ' . We are to prove = arc RS arc R ' S ' . Draw the radii O R , O S , O ' R ' , and O ' S ' . In the ROS and R ' O'S ' RS = R'S ...
Page 81
... subtended by it . B M S Let A B be the chord , and let 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 ...
... subtended by it . B M S Let A B be the chord , and let 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 ...
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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'.