The Contents of the Fifth and Sixth Books of Euclid |
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Page xi
... enunciations no attempt has been made to adhere to Euclid's words . All those propositions which may be viewed either ... enunciation is inverted , as possibly more convenient for reference . This work contains demonstrations of all the ...
... enunciations no attempt has been made to adhere to Euclid's words . All those propositions which may be viewed either ... enunciation is inverted , as possibly more convenient for reference . This work contains demonstrations of all the ...
Page 1
... before entering upon the discussion of the relations between magnitudes . * A point will also be denoted by a capital letter , but this will not lead to any difficulty . H. E. 1 ENUNCIATION . Art . 5. PROPOSITION I. * ( Euc.
... before entering upon the discussion of the relations between magnitudes . * A point will also be denoted by a capital letter , but this will not lead to any difficulty . H. E. 1 ENUNCIATION . Art . 5. PROPOSITION I. * ( Euc.
Page 2
... ENUNCIATION . To prove that ( a + b ) R = aR + bR . Take any rectangle . Draw ( a + b − 1 ) straight lines parallel to one pair of sides , thus dividing it into ( a + b ) compartments . * See Note 1 . In each of these compartments ...
... ENUNCIATION . To prove that ( a + b ) R = aR + bR . Take any rectangle . Draw ( a + b − 1 ) straight lines parallel to one pair of sides , thus dividing it into ( a + b ) compartments . * See Note 1 . In each of these compartments ...
Page 3
... ENUNCIATION . If A > B , then r ( A – B ) = rA − rB . Since let But + A > B , A = B + C. .. rA = r ( B + C ) = rB + rC .. rCrA - rB . CA - B , .. r ( AB ) = rA − rB . - * See Note 1 . [ Prop . 1 . Art . 10. PROPOSITION IV . * ( Euc ...
... ENUNCIATION . If A > B , then r ( A – B ) = rA − rB . Since let But + A > B , A = B + C. .. rA = r ( B + C ) = rB + rC .. rCrA - rB . CA - B , .. r ( AB ) = rA − rB . - * See Note 1 . [ Prop . 1 . Art . 10. PROPOSITION IV . * ( Euc ...
Page 4
... ENUNCIATION . If a > b , prove that ( a - b ) R = aR − bR . Since a > b , and each is a positive integer , .. ( a - b ) is a positive integer which may be called c . .. a = b + c . .. aR = ( b + c ) R = bR + cR . .. cRaRbR .. ( a - b ) ...
... ENUNCIATION . If a > b , prove that ( a - b ) R = aR − bR . Since a > b , and each is a positive integer , .. ( a - b ) is a positive integer which may be called c . .. a = b + c . .. aR = ( b + c ) R = bR + cR . .. cRaRbR .. ( a - b ) ...
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Common terms and phrases
ABCD ABEF angle equal angles reciprocally proportional BCGE BCHD BEFG BEHC bisected BLNO centre circle corresponding sides cross-ratio cutting AC DÊE definition divided drawn perpendicular duplicate ratio ENUNCIATION EQPS equal angles equimultiples Euclid's EXAMPLE exhibited in Fig expressed by Fig fact exhibited four harmonic points four straight lines greater Hence by Prop Hence the triangles hypotenuse integer inversion kind locus mean proportional middle point parallel to BC parallelogram PQRST PROPOSITION Proposition 48 rA rB rA sC radical axis ratio compounded ratio of equality rect rectangle contained relative multiple scale required to prove respectively equal right angle segments shows the fact side BC side corresponding similar figures similar triangles similarly described square on AB square on AC supplementary angles tangents three magnitudes triangle ABC triangle DEF triangles are similar vertex
Popular passages
Page 99 - IF two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals : the triangles shall be equiangular, and shall have those angles equal which are opposite to the homologous sides.
Page xviii - Wherefore, in equal circles &c. QED PROPOSITION B. THEOREM If the vertical angle of a triangle be bisected by a straight line which likewise cuts the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square on the straight line which bisects the angle.
Page 99 - Prove that similar triangles are to one another in the duplicate ratio of their homologous sides.
Page xvi - If the vertical angle of a triangle be bisected by a straight line which also cuts the base, the segments of the base shall have the same ratio which the other sides of the triangle have to one another...
Page 99 - ABC, DEF have one angle in the one equal to one angle in the other, viz. the angle BAC to the angle EDF, and the sides about...
Page xvi - In a right triangle, if a perpendicular is drawn from the vertex of the right angle to the hypotenuse : I.
Page 35 - 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: or, if the multiple of the first be equal to that of the second, the multiple of the third is also equal to that of the fourth : or, if...
Page 84 - If they do not intersect, show that the radical axis is perpendicular to the line joining the centres of the circles...
Page 100 - If an angle of a triangle be bisected by a straight line, which likewise cuts the base; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square of the straight line bisecting the angle.
Page 80 - If four straight lines be proportionals, the rectangle contained by the extremes is equal to the rectangle contained by the means.