Elements of Plane Geometry: For the Use of SchoolsLewis & Sampson, 1844 - 96 σελίδες |
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Αποτελέσματα 1 - 5 από τα 33.
Σελίδα 34
... PROP . III . THEOREM . In the same , or in equal circles , equal angles at ... ( B. I. Def . 23 ) , the point A would fall on D , and the point B on E ; the ... equal angles at the centre are subtended by equal chords . Cor . 2. If the ...
... PROP . III . THEOREM . In the same , or in equal circles , equal angles at ... ( B. I. Def . 23 ) , the point A would fall on D , and the point B on E ; the ... equal angles at the centre are subtended by equal chords . Cor . 2. If the ...
Σελίδα 35
... Prop . III . ) In the same circle , or in equal circles , equal arcs sub- tend equal angles at the centre . Let ( as ... ( B. I. Prop . 22 ) , and the angles ACB , DCE , are equal . Cor . 1. Equal chords subtend equal angles at the centre ...
... Prop . III . ) In the same circle , or in equal circles , equal arcs sub- tend equal angles at the centre . Let ( as ... ( B. I. Prop . 22 ) , and the angles ACB , DCE , are equal . Cor . 1. Equal chords subtend equal angles at the centre ...
Σελίδα 36
... ( Prop . 4 ) ; but these angles are adjacent ; hence each is a right angle . ( B. I. Def . 6. ) PROP . VI . THEOREM ... equal ( Def . 4 ) , therefore AD = DB ( B. I. Prop . 19 , Cor . 2 ) ; hence the two triangles ACD , BCD , having the ...
... ( Prop . 4 ) ; but these angles are adjacent ; hence each is a right angle . ( B. I. Def . 6. ) PROP . VI . THEOREM ... equal ( Def . 4 ) , therefore AD = DB ( B. I. Prop . 19 , Cor . 2 ) ; hence the two triangles ACD , BCD , having the ...
Σελίδα 37
... ( B. I. Prop . 10 , Cor . 3 ) , or else form but one straight line ; but they meet at A , and are therefore not par ... equal distances from the foot of the perpendicular CE , they are equal ( B. I. Prop . 18 ) ; for the same reason AC ...
... ( B. I. Prop . 10 , Cor . 3 ) , or else form but one straight line ; but they meet at A , and are therefore not par ... equal distances from the foot of the perpendicular CE , they are equal ( B. I. Prop . 18 ) ; for the same reason AC ...
Σελίδα 38
... equal to OC , and AH to CK ; hence these triangles are equal ( B. I. Prop . 20 ) , and OH OK . 2d . Let AE be longer than AB ; we have to prove that AB is further from the centre than AE . Draw OI perpendicular to AE . OL is longer than OI ...
... equal to OC , and AH to CK ; hence these triangles are equal ( B. I. Prop . 20 ) , and OH OK . 2d . Let AE be longer than AB ; we have to prove that AB is further from the centre than AE . Draw OI perpendicular to AE . OL is longer than OI ...
Άλλες εκδόσεις - Προβολή όλων
Elements of Plane Geometry: For the Use of Schools - Primary Source Edition Nicholas Tillinghast Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2013 |
Συχνά εμφανιζόμενοι όροι και φράσεις
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
Δημοφιλή αποσπάσματα
Σελίδα 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.
Σελίδα 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.
Σελίδα 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'.
Σελίδα 53 - In any proportion, the product of the means is equal to the product of the extremes.
Σελίδα 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.
Σελίδα 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.
Σελίδα 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.
Σελίδα 59 - The area of a parallelogram is equal to the product of its base and its height: A = bx h.
Σελίδα 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.
Σελίδα 82 - The side of a regular hexagon inscribed in a circle is equal to the radius of the circle.