## The synoptical Euclid; being the first four books of Euclid's Elements of geometry, with exercises, by S.A. Good |

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Resultat 1-5 av 74

Side 10

to AF , the less , and

to AF , the less , and

**join**FC , GB . Because AF is equal ( Constr . ) to AG , and AB ( Hypothesis ) to AC , the two sides FA , AC are equal to the two GA , AB , each to each ; and they contain the angle FAG common to the two triangles ... Side 11

DB equal to AC , the less , and

DB equal to AC , the less , and

**join**DČ . Therefore because in the triangles DBC , ACB , DB is equal to AC , and BC common to both , 1 . The two sides DB , BC , are equal to the two AC , CB , each to each ... Side 12

terminated in the extremity A of the base equal to one another , and likewise their sides CB , DB , that are terminated in B. E c Α . B B

terminated in the extremity A of the base equal to one another , and likewise their sides CB , DB , that are terminated in B. E c Α . B B

**Join**CD . Then , in the case in which the vertex of each of the triangles is without the other ... Side 13

off AE equal to AD ;

off AE equal to AD ;

**join**DE , and upon it , on the side remote from 4 , describe ( I. 1. ) an equilateral triangle DEF ; then**join**AF ; the straight line AF bisects the triangle BAC A D B C Because AD is equal to AE , and AF is common ... Side 14

... at right angles to AB . Take any point D in AC , and ( I. 3. ) make CE equal to CD , and upon DE describe ( I. 1. ) the equilateral triangle DFĒ , and

... at right angles to AB . Take any point D in AC , and ( I. 3. ) make CE equal to CD , and upon DE describe ( I. 1. ) the equilateral triangle DFĒ , and

**join**FC ; the straight line FC drawn from the given point C is 14 EUCLID'S ELEMENTS .### Hva folk mener - Skriv en omtale

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The Synoptical Euclid; Being the First Four Books of Euclid's Elements of ... Uten tilgangsbegrensning - 1854 |

### Vanlige uttrykk og setninger

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

### Populære avsnitt

Side 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...

Side 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.

Side 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.

Side 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.

Side 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.

Side 7 - AB; but things which are equal to the same are equal to one another...

Side 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.

Side 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.

Side 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.

Side 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...