A school Euclid, being books i. & ii. of Euclid's Elements, with notes by C. Mansford1874 |
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Side 22
... angle BAC equal to the angle EDF , the base BC shall be equal to the base EF , and the triangle ABC to the triangle DEF , B and the other angles shall be equal each to each , to which the equal sides are opposite , namely , the angle ...
... angle BAC equal to the angle EDF , the base BC shall be equal to the base EF , and the triangle ABC to the triangle DEF , B and the other angles shall be equal each to each , to which the equal sides are opposite , namely , the angle ...
Side 23
... triangle upon the other . First he shows that AB exactly coincides with DE . Next , that the angle BAC coincides with EDF . Thirdly , that the line AC coincides with DF . And lastly , from these he shows that BC must coincide with EF ...
... triangle upon the other . First he shows that AB exactly coincides with DE . Next , that the angle BAC coincides with EDF . Thirdly , that the line AC coincides with DF . And lastly , from these he shows that BC must coincide with EF ...
Side 28
... angle BAC shall be equal to the angle EDF . A B For if the triangle ABC be applied to the 2 , 3. triangle DEF , so that the point B may be on E , and the straight line BC on the straight line EF , the point C will also coincide with the ...
... angle BAC shall be equal to the angle EDF . A B For if the triangle ABC be applied to the 2 , 3. triangle DEF , so that the point B may be on E , and the straight line BC on the straight line EF , the point C will also coincide with the ...
Side 29
... angle BAG = BGA because BA = BG , and the angle GAC AGC because AC - GC . Therefore the whole angle BAC BGC , that is EDF . Q.E.D. Ex . 1. The diagonal of a rhombus bisects each of the angles through which it passes . Hence show that 2 ...
... angle BAG = BGA because BA = BG , and the angle GAC AGC because AC - GC . Therefore the whole angle BAC BGC , that is EDF . Q.E.D. Ex . 1. The diagonal of a rhombus bisects each of the angles through which it passes . Hence show that 2 ...
Side 30
Euclides Charles Mansford. PROPOSITION 9. PROBLEM . To bisect a given rectilineal angle , that is to divide it into two equal angles . 1 . Let BAC be the given recti- lineal angle . It is required to bisect it . 2 . Take any point D in ...
Euclides Charles Mansford. PROPOSITION 9. PROBLEM . To bisect a given rectilineal angle , that is to divide it into two equal angles . 1 . Let BAC be the given recti- lineal angle . It is required to bisect it . 2 . Take any point D in ...
Andre utgaver - Vis alle
A school Euclid, being books i. & ii. of Euclid's Elements, with notes by C ... Euclides Uten tilgangsbegrensning - 1874 |
A School Euclid. Being Books I. & II. of Euclid's Elements, with Notes ... Euclid,Charles Mansford Uten tilgangsbegrensning - 1875 |
A School Euclid, Being Books I. & II. of Euclid's Elements, with Notes by C ... Euclides Ingen forhåndsvisning tilgjengelig - 2015 |
Vanlige uttrykk og setninger
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 CEA axioms base BC 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 less Let ABC Let the straight obtuse opposite angle opposite sides parallel to CD parallelogram parallelogram ABCD perpendicular produced prop 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