The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago Vitiated These Books, are Corrected and Some of Euclid's Demonstrations are Restored. Also, to this Second Edition is Added the Book of Euclid's Data. In Like Manner Corrected. viz. the first six books, together with the eleventh and twelfth |
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Resultat 1-5 av 20
Side 41
then because each of the angles BAC , BAG is a right angle " , the two straight
lines b . 31.1 . AC , AG upon the opposite G fides of AB , make with it at H the
point A the adjacent angles ecual to two right angles ; therefore CA is in the same
K ...
then because each of the angles BAC , BAG is a right angle " , the two straight
lines b . 31.1 . AC , AG upon the opposite G fides of AB , make with it at H the
point A the adjacent angles ecual to two right angles ; therefore CA is in the same
K ...
Side 129
... K together are any equimultiples of A , and A , C , E together , because if there
be any number of magnitudes equimultiples of as many , each of each , whatever
multiple one of them is of its part , the same multiple is the whole of the wholeb ...
... K together are any equimultiples of A , and A , C , E together , because if there
be any number of magnitudes equimultiples of as many , each of each , whatever
multiple one of them is of its part , the same multiple is the whole of the wholeb ...
Side 167
... make the angle HBL equal to the angle EDK , and the angle BHL equal to the
angle DEK ; therefore the remaining angle at K is equal to the remaining angle at
L. and because the figures ABHG , CDEF are similar , the angle GHB is equal to ...
... make the angle HBL equal to the angle EDK , and the angle BHL equal to the
angle DEK ; therefore the remaining angle at K is equal to the remaining angle at
L. and because the figures ABHG , CDEF are similar , the angle GHB is equal to ...
Side 184
Join BC , CK , and in the circumferences BC , CK take any points X , O , and join
BX , XC , CO , OK . then because in the triangles GBC , GCK the two sides BG ,
GC are equal to the two e . 4. 1. CG , GK , and that they contain equal angles ; the
...
Join BC , CK , and in the circumferences BC , CK take any points X , O , and join
BX , XC , CO , OK . then because in the triangles GBC , GCK the two sides BG ,
GC are equal to the two e . 4. 1. CG , GK , and that they contain equal angles ; the
...
Side 209
... within c . s .. the triangle , or in one of its fides , or without it . First , Let the
center X be within the triangle , and join LX , MX , NX . AB is greater than LX . if
not , AB must either be equal to ; or less than LX ; first , let it be equal . then
because AB ...
... within c . s .. the triangle , or in one of its fides , or without it . First , Let the
center X be within the triangle , and join LX , MX , NX . AB is greater than LX . if
not , AB must either be equal to ; or less than LX ; first , let it be equal . then
because AB ...
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The Elements of Euclid, Viz: The Errors, by which Theon, Or Others, Have ... Robert Simson Uten tilgangsbegrensning - 1775 |
The Elements of Euclid: The Errors by which Theon, Or Others, Have Long ... Robert Simson Uten tilgangsbegrensning - 1827 |
The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago ... Robert Simson Uten tilgangsbegrensning - 1781 |
Vanlige uttrykk og setninger
added alſo altitude angle ABC angle BAC baſe BC is given becauſe Book Book XI caſe circle circle ABCD circumference common cone cylinder Definition demonſtrated deſcribed diameter divided double draw drawn equal equal angles equiangular equimultiples exceſs fame fides firſt folid fore four fourth given angle given in poſition given in ſpecies given magnitude given ratio given ſtraight line greater Greek half join leſs likewiſe magnitude manner meet muſt oppoſite parallel parallelogram perpendicular plane priſm produced PROP proportionals Propoſition pyramid reaſon rectangle remaining right angles ſame ſecond ſegment ſhall ſhewn ſides ſimilar ſolid ſquare ſquare of AC taken THEOR theſe third thro triangle ABC wherefore whole
Populære avsnitt
Side 5 - Let it be granted that a straight line may be drawn from any one point to any other point.
Side 163 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.
Side 48 - If a straight line be divided into any two parts, the squares of the whole line, and of one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square of the other part. Let the straight line AB be divided into any two parts in the point C; the squares of AB, BC are equal to twice the rectangle AB, BC, together with the square of AC.
Side 73 - The straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls without the circle; and no straight line can be drawn from the extremity between that straight line and the circumference, so as not to cut the circle...
Side 105 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz. either the sides adjacent to the equal...
Side 3 - A circle is a plane figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference, are equal to one another.
Side 167 - Similar triangles are to one another in the duplicate ratio of their homologous sides.
Side 54 - AB be the given straight line ; it is required to divide it into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part.
Side 47 - If a straight line be divided into two equal parts, and also into two unequal parts; the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line.
Side 37 - To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.