The first three books of Euclid's Elements of geometry, with theorems and problems, by T. Tate |
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Side 56
DB is therefore greater than DE : But DB is equal to DF ; wherefore DF is greater
than DE , the less than the greater , which is impossible : Therefore the straight ...
But let CD cut AB at right angles ; on also bisects it , that is , AF is equal to FB .
DB is therefore greater than DE : But DB is equal to DF ; wherefore DF is greater
than DE , the less than the greater , which is impossible : Therefore the straight ...
But let CD cut AB at right angles ; on also bisects it , that is , AF is equal to FB .
Side 58
... equal to FB : Also because F is the centre of the circle CDE , CF is equal to FE :
And CF was shown equal to FB ; therefore FE is equal to FB , the less to the
greater , which is impossible ; wherefore F is not the centre of the circles ABC ,
CDE .
... equal to FB : Also because F is the centre of the circle CDE , CF is equal to FE :
And CF was shown equal to FB ; therefore FE is equal to FB , the less to the
greater , which is impossible ; wherefore F is not the centre of the circles ABC ,
CDE .
Side 79
AB in F , and C from the centre F , at the distance FB , describe the semicircle
AHB ; therefore the angle Ahb in a semicircle is ( 111. 31. ) equal to the right
angle at C. But , if the angle c be not a right angle , at the point A , in the straight
line A B ...
AB in F , and C from the centre F , at the distance FB , describe the semicircle
AHB ; therefore the angle Ahb in a semicircle is ( 111. 31. ) equal to the right
angle at C. But , if the angle c be not a right angle , at the point A , in the straight
line A B ...
Side 84
the straight line de touching the circle ABC , find its centre F , and join FE , FB ,
FD ; and then FED is a right angle ( 111. 18. ) : And because DE touches the
circle ABC and DCA cuts it , the rectangle AD , DC is equal ( 111. 36. ) to the
square of ...
the straight line de touching the circle ABC , find its centre F , and join FE , FB ,
FD ; and then FED is a right angle ( 111. 18. ) : And because DE touches the
circle ABC and DCA cuts it , the rectangle AD , DC is equal ( 111. 36. ) to the
square of ...
Side 101
Draw Ds parallel to FB , then BFDs is a parallelogram , and ( E. I. 34. ) Ds is equal
to FB ; and ( constr . ) because the angle sbf is a right angle , therefore ( E. 1. 29. )
the angle DSA is a right angle , that is , Dş is perpendicular to A B , and it has ...
Draw Ds parallel to FB , then BFDs is a parallelogram , and ( E. I. 34. ) Ds is equal
to FB ; and ( constr . ) because the angle sbf is a right angle , therefore ( E. 1. 29. )
the angle DSA is a right angle , that is , Dş is perpendicular to A B , and it has ...
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The first three books of Euclid's Elements of geometry, with theorems and ... Euclides Uten tilgangsbegrensning - 1851 |
The First Three Books of Euclid's Elements of Geometry from the Text of Dr ... Euclid,Thomas Tate Ingen forhåndsvisning tilgjengelig - 2014 |
The First Three Books of Euclid's Elements of Geometry from the Text of Dr ... Euclid,Thomas Tate Ingen forhåndsvisning tilgjengelig - 2014 |
Vanlige uttrykk og setninger
ABCD alternate angle ABC angle ACB angle BAC angle equal base BC BC is equal bisect centre circle ABC circumference coincide common construct demonstrated describe diameter divided double draw equal angles equal to FB equilateral exterior angle extremity figure fore four given point given straight line gnomon greater impossible interior isosceles triangle join less Let ABC likewise line be drawn meet opposite angles opposite sides parallel parallelogram pass perpendicular PROB produced PROP Q. E. D. PROP rectangle contained remaining angle right angles segment semicircle shown sides squares of AC straight line AC Take taken THEOR third touch touches the circle triangle ABC twice the rectangle vertex wherefore whole
Populære avsnitt
Side 6 - If a straight line meets two straight lines, so as to make the two interior angles on the same side of it taken together less than two right angles...
Side 5 - Let it be granted that a straight line may be drawn from any one point to any other point.
Side 20 - If two triangles have two sides of the one equal to two sides of the...
Side 30 - Parallelograms upon equal bases, and between the same parallels, are equal to one another.
Side 17 - Any two angles of a triangle are together less than two right angles. Let ABC be any triangle ; any two of its angles together are less than two right angles.
Side 84 - IF from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it; if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, be equal to the square of the line which meets it, the line which meets it shall touch the circle.
Side 82 - If from any point without a circle two straight lines be drawn, one of -which cuts the circle, and the other touches it; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square of the line which touches it.
Side 11 - UPON the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity.
Side 19 - To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these must be greater than the third, (i.
Side 7 - From the greater of two given straight lines to cut off a part equal to the less. Let AB and C be the two given straight lines, whereof AB is the greater.