First Steps in Geometry: A Series of Hints for the Solution of Geometrical Problems with Notes on Euclid, Useful Working Propositions and Many ExamplesLongmans, Green, and Company, 1887 - 180 σελίδες |
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Αποτελέσματα 1 - 5 από τα 21.
Σελίδα 13
... angles BAH and CA H , we are at once led to notice that our data point to the equality of the triangles HBA and HCA . For we have the angle ... BA , AH equal ANALYSIS AND SYNTHESIS . 13.
... angles BAH and CA H , we are at once led to notice that our data point to the equality of the triangles HBA and HCA . For we have the angle ... BA , AH equal ANALYSIS AND SYNTHESIS . 13.
Σελίδα 14
... BA , AH equal to CA , A H , each to each , and the base BH equal to the base C H. Therefore the angle BAH is equal to the angle CA H. ( Euc . I. , 8. ) It will be noticed that the reasoning from which this solution is obtained is partly ...
... BA , AH equal to CA , A H , each to each , and the base BH equal to the base C H. Therefore the angle BAH is equal to the angle CA H. ( Euc . I. , 8. ) It will be noticed that the reasoning from which this solution is obtained is partly ...
Σελίδα 15
... angle FAH is equal to the angle G A H. It is probable , however , that the geometrician , being led in this way to ... B A , A H to C A , A H , respectively . Thus we gather an important rule . Having tracked out , analytically or ...
... angle FAH is equal to the angle G A H. It is probable , however , that the geometrician , being led in this way to ... B A , A H to C A , A H , respectively . Thus we gather an important rule . Having tracked out , analytically or ...
Σελίδα 19
... angle But there seems no obvious We have AF method of proving this relation . Let us consider our construction . equal to A C. But AC is equal to AB . Thus BA ... angle . ( Euc . III . , 31. ) We therefore join CF , and note that it is ...
... angle But there seems no obvious We have AF method of proving this relation . Let us consider our construction . equal to A C. But AC is equal to AB . Thus BA ... angle . ( Euc . III . , 31. ) We therefore join CF , and note that it is ...
Σελίδα 21
... angle DFG equal to the angle D G F ( Euc . I. , 5 ) ; therefore the angle B F G greater than the angle BG F ; and BG greater than BF ( Euc . I. , 19 ) ; that is , BD , DC , together greater than B A , A C together . Or , we might notice ...
... angle DFG equal to the angle D G F ( Euc . I. , 5 ) ; therefore the angle B F G greater than the angle BG F ; and BG greater than BF ( Euc . I. , 19 ) ; that is , BD , DC , together greater than B A , A C together . Or , we might notice ...
Άλλες εκδόσεις - Προβολή όλων
First Steps in Geometry: A Series of Hints for the Solution of Geometrical ... Richard Anthony Proctor Προβολή αποσπασμάτων - 1888 |
First Steps in Geometry: A Series of Hints for the Solution of Geometrical ... Richard Anthony Proctor Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2018 |
First Steps in Geometry: A Series of Hints for the Solution of Geometrical ... Richard Anthony Proctor Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2013 |
Συχνά εμφανιζόμενοι όροι και φράσεις
A B C D A B is equal ABCD angle ABC angle BA angle equal base bisector bisects the angle circle diagonals equal and parallel equal angles equal sides equal to DC equal to half equal to twice Euclid exterior angles given angle given line given point given straight line greater Hence hypotenuse intersect isosceles triangle KHGE lines bisecting lines drawn locus maxima and minima obtuse opposite sides parallelogram perpendicular point F problem produced proof Prop proposition quadrilateral R. A. PROCTOR rect rectangle A C rectangle contained respects Euc rhombus right angles right-angled triangle sides A B solution square on CD squares on A C straight line A B theorems trapezium triangle ABC twice the rectangle vertex vertical angle
Δημοφιλή αποσπάσματα
Σελίδα 80 - If two triangles have two sides of the one equal to two sides of the...
Σελίδα 144 - To draw a straight line at right angles to a given straight line, from a given point in the same.
Σελίδα 175 - IF a straight line be divided into two equal, and also into two unequal parts ; the squares of the two unequal parts are together double of the square of half the line, and of the square of the line between the points of section.
Σελίδα 136 - AB into two parts, so that the rectangle contained by the whole line and one of the parts, shall be equal to the square on the other part.