Euclidian Geometry |
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Side 14
... of these lines produced would fall along the former position of the other . Thus
the arms of A would fall along the former positions of the arms of B ; . . the angles
A , B are equal to one another . Similarly the angles H , K may be proved equal .
... of these lines produced would fall along the former position of the other . Thus
the arms of A would fall along the former positions of the arms of B ; . . the angles
A , B are equal to one another . Similarly the angles H , K may be proved equal .
Side 21
ARF would be a straight line ; ( 1 . 10 ) and thus two straight lines would inclose a
space , which is impossible . . . AR is not I to BC . Similarly it may be proved that
no other straight line besides AQ , drawn from A to BC , is I to it . INEQUALITIES .
ARF would be a straight line ; ( 1 . 10 ) and thus two straight lines would inclose a
space , which is impossible . . . AR is not I to BC . Similarly it may be proved that
no other straight line besides AQ , drawn from A to BC , is I to it . INEQUALITIES .
Side 36
13 ) but it is also equal to it , which is impossible . . . PQ , RS do not meet when
produced towards Q , S . Similarly it may be proved that they will not meet if
produced towards P , R . . . they are parallel . Hence are easily deduced the
following ...
13 ) but it is also equal to it , which is impossible . . . PQ , RS do not meet when
produced towards Q , S . Similarly it may be proved that they will not meet if
produced towards P , R . . . they are parallel . Hence are easily deduced the
following ...
Side 57
... be proved that the O QC is = the square on RQ ; . . the whole figure PHNQ , that
is the square on PQ , is = the squares on PR , RQ . PROPOSITION XXXVI . If the
square described on one of AREAS .
... be proved that the O QC is = the square on RQ ; . . the whole figure PHNQ , that
is the square on PQ , is = the squares on PR , RQ . PROPOSITION XXXVI . If the
square described on one of AREAS .
Side 80
Euclid proves the case in which the given equal sides are opposite to equal
angles in each thus : Let ABC , DEF be two as having ABC = - DEF , - ACB = DFE
, and side AB = side DE . Then BC and EF are equal to one another , for , if not ,
one ...
Euclid proves the case in which the given equal sides are opposite to equal
angles in each thus : Let ABC , DEF be two as having ABC = - DEF , - ACB = DFE
, and side AB = side DE . Then BC and EF are equal to one another , for , if not ,
one ...
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base bisect Cambridge centre chapter chord circle circumference cloth College common complete construction contained Crown 8vo DEFINITION describe diameter difference divided double draw drawn Edition elementary English equal equiangular Examination Examples extremities fall fcap figure former four GEOMETRY given point given straight line Grammar greater Hence illustrations impossible inscribed introduction Join language Latin less magnitudes Master Mathematical meet method Notes opposite parallel parallelogram pass perimeter perpendicular plane polygon possible present principles PROBLEM produced Professor proportional PROPOSITION prove ratio rect rectangle render respectively revised right angles Schools segment selected sides similar Similarly square taken tangent THEOREM third touch TREATISE triangle twice University volume whole
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Side 157 - The first of four magnitudes is said to have the same ratio to the second, which the third has to the fourth, when any equimultiples whatsoever of the first and third being taken, and any equimultiples whatsoever of the second and fourth ; if the multiple of the first be less than that of the second, the multiple of the third is also less than that of the fourth...
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