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 67
Side 6
... A B and BC ( Fig . 2 ) , and taking P as the given point , to draw PDE , so that PD and DE might be as nearly equal as possible . The proper way , however , is to draw a straight line , P E , bisect it in D , and through the points D E ...
... A B and BC ( Fig . 2 ) , and taking P as the given point , to draw PDE , so that PD and DE might be as nearly equal as possible . The proper way , however , is to draw a straight line , P E , bisect it in D , and through the points D E ...
Side 8
... equal angles with any side . It will be noticed , also , that neither diagonal bisects the other . If we had a ... A B , BC , CD , and D A. It is sometimes convenient to draw a part of the figure in darker lines than the rest . We may ...
... equal angles with any side . It will be noticed , also , that neither diagonal bisects the other . If we had a ... A B , BC , CD , and D A. It is sometimes convenient to draw a part of the figure in darker lines than the rest . We may ...
Side 11
... equal to E B. Now , we cannot fail to see that the data involve the equality of the triangles ACE , CE B , by Euc ... BC , CE , the angles ACE and BCE will be equal ( Euc . I. , 8 ) ; but these angles are equal , being right angles ...
... equal to E B. Now , we cannot fail to see that the data involve the equality of the triangles ACE , CE B , by Euc ... BC , CE , the angles ACE and BCE will be equal ( Euc . I. , 8 ) ; but these angles are equal , being right angles ...
Side 12
... equal , respectively , to B C , C E , the triangles ACE , BC E are equal in all respects ; therefore the angle CA E will be equal to the angle CBE . Now , these angles are equal ; therefore we might expect the reversal of the process to ...
... equal , respectively , to B C , C E , the triangles ACE , BC E are equal in all respects ; therefore the angle CA E will be equal to the angle CBE . Now , these angles are equal ; therefore we might expect the reversal of the process to ...
Side 16
... BC , and we know from Euc . I. , 19 , that if DC is greater than B C , then ... equal to the angle ACB ( Euc . I. , 5 ) , and two angles of a triangle being ... BC . We will next try a proposition slightly more difficult . 16 GEOMETRICAL ...
... BC , and we know from Euc . I. , 19 , that if DC is greater than B C , then ... equal to the angle ACB ( Euc . I. , 5 ) , and two angles of a triangle being ... BC . We will next try a proposition slightly more difficult . 16 GEOMETRICAL ...
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First Steps in Geometry: A Series of Hints for the Solution of Geometrical ... Richard Anthony Proctor Ingen forhåndsvisning tilgjengelig - 2018 |
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
Side 80 - If two triangles have two sides of the one equal to two sides of the...
Side 144 - To draw a straight line at right angles to a given straight line, from a given point in the same.
Side 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.
Side 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.