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 sider |
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Resultat 1-5 av 15
Side 19
... diameter of this circle , B C F is a right angle . ( Euc . III . , 31. ) We therefore join CF , and note that it is perpendicular to BC.1 Here is an instance of the advantage of carefully constructed figures . The relation arrived at by ...
... diameter of this circle , B C F is a right angle . ( Euc . III . , 31. ) We therefore join CF , and note that it is perpendicular to BC.1 Here is an instance of the advantage of carefully constructed figures . The relation arrived at by ...
Side 24
... diameters of these equal circles , are severally double of A B. 2 In constructing the figure , proceed thus : -Take A D equal to A E , and join D E ; then take P , a point dividing D E into unequal parts . Prop . 5. This gives us the ...
... diameters of these equal circles , are severally double of A B. 2 In constructing the figure , proceed thus : -Take A D equal to A E , and join D E ; then take P , a point dividing D E into unequal parts . Prop . 5. This gives us the ...
Side 52
... diameter APEC F FIG . 27 . E FIG . 28 . through P. Its bisection , E , is the centre of the circle . This is one point of the required locus . Draw next the chord BPD at right angles to A C. Then the point P is itself the bisection of ...
... diameter APEC F FIG . 27 . E FIG . 28 . through P. Its bisection , E , is the centre of the circle . This is one point of the required locus . Draw next the chord BPD at right angles to A C. Then the point P is itself the bisection of ...
Side 53
... diameter , and we look for a proof that a chord drawn as F PHG would be bisected in H , where it meets the circle thus drawn . It will clearly be well to join EH . When this is done , one of two well - known pro- perties can hardly fail ...
... diameter , and we look for a proof that a chord drawn as F PHG would be bisected in H , where it meets the circle thus drawn . It will clearly be well to join EH . When this is done , one of two well - known pro- perties can hardly fail ...
Side 54
... diameter ( Euc . III . , 31 ) . But FG is any chord through P. Therefore the bisections of all such chords lie on the circle EHP . Also it is clear that every point on this circle bisects some chord through P. Therefore this circle is ...
... diameter ( Euc . III . , 31 ) . But FG is any chord through P. Therefore the bisections of all such chords lie on the circle EHP . Also it is clear that every point on this circle bisects some chord through P. Therefore this circle is ...
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First Steps in Geometry: A Series of Hints for the Solution of Geometrical ... Richard Anthony Proctor Ingen forhåndsvisning tilgjengelig - 2013 |
Vanlige uttrykk og setninger
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
Populære avsnitt
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