The synoptical Euclid; being the first four books of Euclid's Elements of geometry, with exercises, by S.A. Good |
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Αποτελέσματα 1 - 5 από τα 25.
Σελίδα 23
The exterior angle BDC of the triangle CDE is greater 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 exterior angle BDC of the triangle CDE is greater 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.
Σελίδα 35
to BC ; for the same reason , EF is equal to BC ; wherefore ( Ax . 1. ) 1. AD is equal to EF ; The parallelograms ABCD , DBCF , are equal to one and DE is common ; therefore ( Ax . 2. or 3. ) 2. The whole , or the remainder , AE is ...
to BC ; for the same reason , EF is equal to BC ; wherefore ( Ax . 1. ) 1. AD is equal to EF ; The parallelograms ABCD , DBCF , are equal to one and DE is common ; therefore ( Ax . 2. or 3. ) 2. The whole , or the remainder , AE is ...
Σελίδα 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 and ...
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 and ...
Σελίδα 40
... KD , the other parallelograms which make up the whole figure ABCD , which are therefore called the complements . The complement BK is equal to the complement KD . and for the same reason , E B H G 1 . Because ABCD is a parallelogram ...
... KD , the other parallelograms which make up the whole figure ABCD , which are therefore called the complements . The complement BK is equal to the complement KD . and for the same reason , E B H G 1 . Because ABCD is a parallelogram ...
Σελίδα 44
Let ABC be a right - angled triangle having the right angle BAC ; the square described upon the side BC is equal to the squares described upon BA , AC . G · 1 . For the same reason B H D L E On BC describe ( I. 46. ) ...
Let ABC be a right - angled triangle having the right angle BAC ; the square described upon the side BC is equal to the squares described upon BA , AC . G · 1 . For the same reason B H D L E On BC describe ( I. 46. ) ...
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ABCD AC is equal AF is equal angle ABC angle ACB angle BAC angle BCD angle equal base BC bisected centre circle ABC circumference coincide common demonstrated describe diameter distance divided double draw equal angles equal Constr exterior angle extremity fall figure four given circle given point given straight line given triangle greater impossible inscribed join less Let ABC likewise manner meet opposite angles parallel parallelogram pass pentagon perpendicular point F produced Q.E.D. PROP reason rectangle contained rectilineal figure remaining angle required to describe right angles segment semicircle shown side BC sides square of AC straight line AC touches the circle triangle ABC twice the rectangle wherefore whole
Δημοφιλή αποσπάσματα
Σελίδα 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...
Σελίδα 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.
Σελίδα 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.
Σελίδα 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.
Σελίδα 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.
Σελίδα 7 - AB; but things which are equal to the same are equal to one another...
Σελίδα 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.
Σελίδα 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.
Σελίδα 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.
Σελίδα 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...
Πληροφορίες βιβλιογραφίας
| Τίτλος | The synoptical Euclid; being the first four books of Euclid's Elements of geometry, with exercises, by S.A. Good |
| Συγγραφέας | Euclides |
| Επιμελητής | Samuel A Good |
| Εκδόθηκε | 1853 |
| Πρωτότυπο από | Πανεπιστήμιο Οξφόρδης |
| Ψηφιοποιήθηκε στις | 7 Σεπτ. 2006 |
| Εξαγωγή αναφοράς | BiBTeX EndNote RefMan |