A School Euclid. Being Books I.&II. of Euclid's Elements. With Notes, Exercises and Explanations ... By C. Mansford1874 |
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Página 21
... 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 , to cut off a part equal to the less ...
... 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 , to cut off a part equal to the less ...
Página 24
... 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 , AGB ; therefore ...
... 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 , AGB ; therefore ...
Página 25
... the less , [ I. 3. ] 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.
... the less , [ I. 3. ] 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 ...
Otras ediciones - Ver todas
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
AC is equal adjacent angles alternate angles angle ABC angle ACB angle AGH angle BAC angle BCD angle EDF angle equal angles are equal angles CBA axioms base BC BC is equal bisect centre circle coincide Const diagonals diameter double equal sides equal to BC equal to twice equilateral triangle Euclid exterior angle fore four right angles given point given rectilineal angle given straight line gnomon half a right hypotenuse interior and opposite isosceles triangle join Let ABC Let the straight obtuse opposite angle opposite sides parallel to CD parallelogram parallelogram ABCD perpendicular produced prop PROPOSITION quadrilateral rectangle AC rectangle contained remaining angle rhombus right angles right-angled triangle shew side BC sides equal square described square on AC THEOREM third angle triangle ABC triangle DEF truths twice the rectangle unequal