A School Euclid. Being Books I.&II. of Euclid's Elements. With Notes, Exercises and Explanations ... By C. Mansford |
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Página 21
Construct and demonstrate this problem by joining A with C instead of with B. Draw the figure of this proposition when the given point is in the given line produced . PROPOSITION 3. PROBLEM . From the greater of two given straight lines ...
Construct and demonstrate this problem by joining A with C instead of with B. Draw the figure of this proposition when the given point is in the given line produced . PROPOSITION 3. PROBLEM . From the greater of two given straight lines ...
Página 24
and join FC , GB . Because A Fis equal 3. to AG , [ Con . ] and AB to AC , [ Hyp . ] the two sides FA , AC are equal to the two sides GA , AB , each to each ; and they contain the angle FAG common to the two triangles AFC ...
and join FC , GB . Because A Fis equal 3. to AG , [ Con . ] and AB to AC , [ Hyp . ] the two sides FA , AC are equal to the two sides GA , AB , each to each ; and they contain the angle FAG common to the two triangles AFC ...
Página 25
and join DC . B 3 . Then , because in the triangles DBC , ACB , DB is equal to AC , [ Con . ] and BC is common to both , the two sides DB , BC are equal to the two sides AC , CB , each to each ; and the BOOK I. 6 . 25.
and join DC . B 3 . Then , because in the triangles DBC , ACB , DB is equal to AC , [ Con . ] and BC is common to both , the two sides DB , BC are equal to the two sides AC , CB , each to each ; and the BOOK I. 6 . 25.
Página 27
Join CD . Then in the case in which the vertex of each triangle is without the other triangle ; because AC is equal to AD [ Hyp . ] , the angle ACD is equal to the angle ADC . [ I. 5. ] But the angle ACD is greater than the angle BCD ...
Join CD . Then in the case in which the vertex of each triangle is without the other triangle ; because AC is equal to AD [ Hyp . ] , the angle ACD is equal to the angle ADC . [ I. 5. ] But the angle ACD is greater than the angle BCD ...
Página 29
Join AG . Then the angle BAG = BGA because BA = BG , and the angle GAC = AGC because AC - GC . Therefore the whole anglo BAC - BGC , that is EDF . Q.E.D. Ex . 1. The diagonal of a rhombus bisects each of the angles through which it ...
Join AG . Then the angle BAG = BGA because BA = BG , and the angle GAC = AGC because AC - GC . Therefore the whole anglo BAC - BGC , that is EDF . Q.E.D. Ex . 1. The diagonal of a rhombus bisects each of the angles through which it ...
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A School Euclid. Being Books I.&II. of Euclid's Elements. With Notes ... Euclides Vista completa - 1874 |
A School Euclid, Being Books I. & II. of Euclid's Elements, with Notes by C ... Euclides Sin vista previa disponible - 2015 |
Términos y frases comunes
ABCD AC is equal alternate angles angle ABC angle ACB angle BAC angle BCD angle EDF angle equal apply axioms base BC bisect BOOK called centre circle coincide common Const construction definitions demonstration describe diagonals diameter difference divided double draw equal sides equilateral triangle Euclid exercise exterior angle fall figure fore geometry given point given rectilineal given straight line gnomon greater Hence isosceles triangle join length less Let ABC meet method namely opposite angle opposite sides parallel parallel to CD parallelogram perpendicular PROBLEM produced prop PROPOSITION proved quadrilateral reason rectangle contained rectilineal figure remainder result right angles side BC sides square on AC Take THEOREM things triangle ABC true truths twice the rectangle unequal units whole
Información bibliográfica
| Título | A School Euclid. Being Books I.&II. of Euclid's Elements. With Notes, Exercises and Explanations ... By C. Mansford |
| Autor | Euclides |
| Editor | Charles MANSFORD |
| Publicado | 1874 |
| Procedencia del original | Biblioteca Británica |
| Digitalizado | 27 Nov. 2020 |
| Exportar cita | BiBTeX EndNote RefMan |