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 27
... bisector of the angle QCE . The only property con- nected with the bisector of an angle which seems likely to help us is this one , that the bisector of the vertical Then there is no risk that accidental relations will appear as ...
... bisector of the angle QCE . The only property con- nected with the bisector of an angle which seems likely to help us is this one , that the bisector of the vertical Then there is no risk that accidental relations will appear as ...
Side 36
... bisector of the angle contained by the equal sides . Hence C E is at right angles to QL and bisects QL in E. It is a very obvious con- sideration , at this point , that if we join C , L we shall have QC , L an isosceles triangle , C , Q ...
... bisector of the angle contained by the equal sides . Hence C E is at right angles to QL and bisects QL in E. It is a very obvious con- sideration , at this point , that if we join C , L we shall have QC , L an isosceles triangle , C , Q ...
Side 49
... bisector of the angle KAL . It is clear , also , that every point in the bisector of the angle KAL fulfils the required conditions . For , let Q be such a point , and draw SQR at right angles to AQ ; then the triangles A Q S and A QR ...
... bisector of the angle KAL . It is clear , also , that every point in the bisector of the angle KAL fulfils the required conditions . For , let Q be such a point , and draw SQR at right angles to AQ ; then the triangles A Q S and A QR ...
Side 68
... bisector , and drawing perpendiculars FH and FG on A B and A C respec- tively , we see that FH is equal to FG ( since the triangles FAH , FAG are equal in all respects , Euc . I. , 26 ) . We describe a circle H G K , with centre F and ...
... bisector , and drawing perpendiculars FH and FG on A B and A C respec- tively , we see that FH is equal to FG ( since the triangles FAH , FAG are equal in all respects , Euc . I. , 26 ) . We describe a circle H G K , with centre F and ...
Side 70
... bisector of the angle BAC and can be drawn at once . RDO PQ at right angles to AE can also be drawn , and gives P , a point on the required circle ; for DO - OP . Then one can hardly miss the relation that the square on RN is equal to ...
... bisector of the angle BAC and can be drawn at once . RDO PQ at right angles to AE can also be drawn , and gives P , a point on the required circle ; for DO - OP . Then one can hardly miss the relation that the square on RN is equal to ...
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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
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.