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 páginas |
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Página 11
... respects , and thence to the equality of A E , E B. In the analytical solution we should argue thus : - We have to show that A E is equal to E B. Now , if AE is equal to EB , then since AC , CE are respectively equal to BC , CE , the ...
... respects , and thence to the equality of A E , E B. In the analytical solution we should argue thus : - We have to show that A E is equal to E B. Now , if AE is equal to EB , then since AC , CE are respectively equal to BC , CE , the ...
Página 12
... 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 lead at once to the solution of our problem . This , however , is not the case ...
... 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 lead at once to the solution of our problem . This , however , is not the case ...
Página 20
... respects . ( Euc . I. , 4. ) we would show that it is not accidental and suffice to exhibit its cause . 1 This problem is not explicitly stated in Euclid . It is con- tained implicitly in Bk . I. , Props . 10-12 . It should be included ...
... respects . ( Euc . I. , 4. ) we would show that it is not accidental and suffice to exhibit its cause . 1 This problem is not explicitly stated in Euclid . It is con- tained implicitly in Bk . I. , Props . 10-12 . It should be included ...
Página 21
... . This is easily done . The student sees at once that the proof involves the equality ( in all respects ) of the triangles AHF , AH C. He had the angle A F H equal to the angle ACH , AF equal to A C , and AH THEOREMS . 21.
... . This is easily done . The student sees at once that the proof involves the equality ( in all respects ) of the triangles AHF , AH C. He had the angle A F H equal to the angle ACH , AF equal to A C , and AH THEOREMS . 21.
Página 22
... respects ( Euc . I. , 4 ) . Hence FH is equal to HC , and the angles at H are right angles . Thus the triangles D HF and DHC are equal in all respects ( Euc . I. , 4 ) . Therefore DF is equal to DC . But BD and DF are together greater ...
... respects ( Euc . I. , 4 ) . Hence FH is equal to HC , and the angles at H are right angles . Thus the triangles D HF and DHC are equal in all respects ( Euc . I. , 4 ) . Therefore DF is equal to DC . But BD and DF are together greater ...
Otras ediciones - Ver todas
First Steps in Geometry: A Series of Hints for the Solution of Geometrical ... Richard Anthony Proctor Vista de fragmentos - 1888 |
First Steps in Geometry: A Series of Hints for the Solution of Geometrical ... Richard Anthony Proctor Sin vista previa disponible - 2018 |
First Steps in Geometry: A Series of Hints for the Solution of Geometrical ... Richard Anthony Proctor Sin vista previa disponible - 2013 |
Términos y frases comunes
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
Pasajes populares
Página 80 - If two triangles have two sides of the one equal to two sides of the...
Página 144 - To draw a straight line at right angles to a given straight line, from a given point in the same.
Página 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.
Página 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.