Elements of Plane Geometry: For the Use of SchoolsLewis & Sampson, 1844 - 96 sider |
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Side 34
... subtended by equal chords . Cor . 2. If the angle at the centre be bisected , both the arc and chord subtending it are also bisected . Cor . 3. If the angles are unequal the chords will be unequal . Scholium . The above reasoning ...
... subtended by equal chords . Cor . 2. If the angle at the centre be bisected , both the arc and chord subtending it are also bisected . Cor . 3. If the angles are unequal the chords will be unequal . Scholium . The above reasoning ...
Side 35
... subtended by equal chords . Cor . 3. If the chords are unequal the angles and arcs which they subtend are unequal . PROP . V. THEOREM . An angle at the centre , subtended by half a semi - cir- cumference , is a right angle . Let the ...
... subtended by equal chords . Cor . 3. If the chords are unequal the angles and arcs which they subtend are unequal . PROP . V. THEOREM . An angle at the centre , subtended by half a semi - cir- cumference , is a right angle . Let the ...
Side 36
... subtends . PROP . VII . THEOREM . Through three given points , not in the same line , one circumference may be made to pass , and only one . To prove this , let A , B , D , be the given points , and 7 BOOK II . ] DISTANCE OF CHORDS FROM ...
... subtends . PROP . VII . THEOREM . Through three given points , not in the same line , one circumference may be made to pass , and only one . To prove this , let A , B , D , be the given points , and 7 BOOK II . ] DISTANCE OF CHORDS FROM ...
Side 41
... subtending the same arc . Cor . 2. Inscribed angles subtended by the same or equal arcs are equal , because they are each equal to the same angle at the centre . Cor . 3. An inscribed angle , which is subtended by a semi - circumference ...
... subtending the same arc . Cor . 2. Inscribed angles subtended by the same or equal arcs are equal , because they are each equal to the same angle at the centre . Cor . 3. An inscribed angle , which is subtended by a semi - circumference ...
Side 50
... subtended by a semi - circumference ( B. II . Prop . 9 , and B. II . Prop . 11 , Cor . 3 ) . Cor . The right - angled triangles BDC , BAC , having CB common and CD = CA , are equal ( B. I. Prop . 20 ) , and hence BD - BA ; therefore two ...
... subtended by a semi - circumference ( B. II . Prop . 9 , and B. II . Prop . 11 , Cor . 3 ) . Cor . The right - angled triangles BDC , BAC , having CB common and CD = CA , are equal ( B. I. Prop . 20 ) , and hence BD - BA ; therefore two ...
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Elements of Plane Geometry: For the Use of Schools Nicholas Tillinghast Uten tilgangsbegrensning - 1844 |
Elements of Plane Geometry: For the Use of Schools Nicholas Tillinghast Uten tilgangsbegrensning - 1844 |
Elements of Plane Geometry: For the Use of Schools - Primary Source Edition Nicholas Tillinghast Ingen forhåndsvisning tilgjengelig - 2013 |
Vanlige uttrykk og setninger
ABCD adjacent angles allel alternate angles altitude angle ABC angles ABD angles is equal antecedent and consequent B. I. Ax base centre circle whose radius circumference circumscribed circumscribed circle Converse of Prop describe an arc diagonal diameter divide draw the line equal angles equal B. I. Prop equal chords equal Prop equal respectively equiangular equivalent feet given angle given line given point given side half hence the triangles hypotenuse included angle inscribed angle Let the triangles line drawn linear units longer than AC multiplied number of sides oblique lines parallel to CD parallelogram perimeter perpendicular PROBLEM prove radii rectangle regular polygons respectively equal right angles Prop right-angled triangle Scholium sides AC similar subtended tangent THEOREM three sides triangles ABC triangles are equal vertex
Populære avsnitt
Side 31 - A circle is a plane figure bounded by a curved line, every point of which is equally distant from a point within called the center.
Side 63 - The square described on the hypothenuse of a right-angled triangle is equivalent to the sum of the squares described on the other two sides.
Side 70 - 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'.
Side 53 - In any proportion, the product of the means is equal to the product of the extremes.
Side 87 - 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.
Side 54 - In a series of equal ratios, any antecedent is to its consequent, as the sum of all the antecedents is to the sum of all the consequents. Let a: 6 = c: d = e :/. Then, by Art.
Side 81 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Side 59 - The area of a parallelogram is equal to the product of its base and its height: A = bx h.
Side 61 - From this proposition it is evident, that the square described on the difference of two lines is equivalent to the sum of the squares described on the lines respectively, minus twice the rectangle contained by the lines.
Side 82 - The side of a regular hexagon inscribed in a circle is equal to the radius of the circle.