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 31
Side 19
... perpendicular to BC.1 Here is an instance of the advantage of carefully constructed figures . The relation arrived at by a tolerably obvious line of reasoning might be overlooked for awhile . But if the figure has been constructed ...
... perpendicular to BC.1 Here is an instance of the advantage of carefully constructed figures . The relation arrived at by a tolerably obvious line of reasoning might be overlooked for awhile . But if the figure has been constructed ...
Side 25
... perpendicular from the vertex on the base of an isosceles triangle bisects the base and also the vertical angle . Draw B P M FIG . 12 . A M perpendicular to the assumed line DE ; then the angle M AD is equal to the angle M A E , and ...
... perpendicular from the vertex on the base of an isosceles triangle bisects the base and also the vertical angle . Draw B P M FIG . 12 . A M perpendicular to the assumed line DE ; then the angle M AD is equal to the angle M A E , and ...
Side 27
... likely to help us is this one , that the bisector of the vertical Then there is no risk that accidental relations will appear as necessary ones . angle of an isosceles triangle is perpendicular to and bisects PROBLEMS . 27.
... likely to help us is this one , that the bisector of the vertical Then there is no risk that accidental relations will appear as necessary ones . angle of an isosceles triangle is perpendicular to and bisects PROBLEMS . 27.
Side 28
... perpendicular to and bisects the base . Now , we can make an isosceles triangle of which C shall be the vertex and CQ a side , for we have only to take CE equal to CQ , and to join Q E , cutting C B in F. Then , by the property just ...
... perpendicular to and bisects the base . Now , we can make an isosceles triangle of which C shall be the vertex and CQ a side , for we have only to take CE equal to CQ , and to join Q E , cutting C B in F. Then , by the property just ...
Side 41
... perpendicular to A B C is less than B F. This is evident ; for in the right - angled triangle B DG , the angle D BG is less than a right angle ; therefore DG is less than B D , —that is , than BF . VIII . PERIMETERS OF TRIANGLES . Let ...
... perpendicular to A B C is less than B F. This is evident ; for in the right - angled triangle B DG , the angle D BG is less than a right angle ; therefore DG is less than B D , —that is , than BF . VIII . PERIMETERS OF TRIANGLES . Let ...
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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.