The synoptical Euclid; being the first four books of Euclid's Elements of geometry, with exercises, by S.A. Good |
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Página 23
For the same reason , 2 . The exterior angle CEB of the triangle ABE is greater than BAC ; and it has been demonstrated that the angle BDC ' is greater than CEB ; much more then 3 . The angle BDC is greater than the angle BAC .
For the same reason , 2 . The exterior angle CEB of the triangle ABE is greater than BAC ; and it has been demonstrated that the angle BDC ' is greater than CEB ; much more then 3 . The angle BDC is greater than the angle BAC .
Página 35
to BC ; for the same reason , EF is equal to BC ; wherefore ( Ax . 1. ) 1 . AD is equal to EF ; and DE is common ; therefore ( Ax . 2. or 3. ) 2 . The whole , or the remainder , AE is equal to the whole , or the remainder , DF ; AB also ...
to BC ; for the same reason , EF is equal to BC ; wherefore ( Ax . 1. ) 1 . AD is equal to EF ; and DE is common ; therefore ( Ax . 2. or 3. ) 2 . The whole , or the remainder , AE is equal to the whole , or the remainder , DF ; AB also ...
Página 36
For the like reason 5 . The parallelogram EFGH is equal to the same EBCH . Therefore also 6 . The parallelogram ABCD is equal to EFGH . Wherefore , parallelograms , & c . Q.E.D. PROP . XXXVII . - THEOREM . Triangles upon the same base ...
For the like reason 5 . The parallelogram EFGH is equal to the same EBCH . Therefore also 6 . The parallelogram ABCD is equal to EFGH . Wherefore , parallelograms , & c . Q.E.D. PROP . XXXVII . - THEOREM . Triangles upon the same base ...
Página 40
The triangle AEK is equal to the triangle AHK ; and for the same reason , 3 . The triangle KGC is equal to the triangle KFC . Then because the triangle AEK is equal to the triangle AHK , and the triangle KGC to KFC ; ( Ax . 2. ) 4 .
The triangle AEK is equal to the triangle AHK ; and for the same reason , 3 . The triangle KGC is equal to the triangle KFC . Then because the triangle AEK is equal to the triangle AHK , and the triangle KGC to KFC ; ( Ax . 2. ) 4 .
Página 44
For the same reason 2 . AB and AH are in the same straight line . And because the angle DBC is equal to the angle FBA , each of them being a right angle , add to each the angle ABC , and ( Ax . 2. ) 3 . The whole angle DBA is equal to ...
For the same reason 2 . AB and AH are in the same straight line . And because the angle DBC is equal to the angle FBA , each of them being a right angle , add to each the angle ABC , and ( Ax . 2. ) 3 . The whole angle DBA is equal to ...
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ABCD AC is equal AF is equal angle ABC angle ACB angle BAC angle BCD angle equal base base BC bisected centre circle ABC circumference coincide common demonstrated describe diameter distance divided double draw equal angles exterior angle extremity fall figure four given circle given point given straight line given triangle gnomon greater impossible inscribed join less Let ABC likewise manner meet opposite angles parallel parallelogram pass pentagon perpendicular point F PROBLEM produced Q.E.D. PROP reason rectangle contained rectilineal figure remaining angle required to describe right angles segment semicircle shown sides square of AC straight line AC THEOREM touches the circle triangle ABC twice the rectangle wherefore whole
Pasajes populares
Página 26 - 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...
Página 22 - If from the ends of a side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle.
Página 32 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Página 1 - A plane superficies is that in which any two points being taken, the straight line between them lies wholly in that superficies. vm. "A plane angle is the inclination of two lines to one another in a plane, which meet together, but are not in the same direction.
Página 97 - 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.
Página 7 - AB; but things which are equal to the same are equal to one another...
Página 14 - To bisect a given finite straight line, that is, to divide it into two equal parts. Let AB be the given straight line : it is required to divide it intotwo equal parts.
Página 53 - 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.
Página 41 - To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.
Página 52 - If a straight line be bisected, and produced to any point; the rectangle contained by the whole line thus produced, and the part of it produced...
Información bibliográfica
| Título | The synoptical Euclid; being the first four books of Euclid's Elements of geometry, with exercises, by S.A. Good |
| Autor | Euclides |
| Editor | Samuel A Good |
| Publicado | 1853 |
| Procedencia del original | Universidad de Oxford |
| Digitalizado | 7 Sep. 2006 |
| Exportar cita | BiBTeX EndNote RefMan |