## 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 31

Side 10

А B В F G D E In BD take any point F , and from AE , the greater , cut off AG equal ( I. 3. ) to AF , the less , and join FC , GB . Because

А B В F G D E In BD take any point F , and from AE , the greater , cut off AG equal ( I. 3. ) 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 ... Side 13

Let ABC , DEF , be two triangles having the two sides AB , AC ,

Let ABC , DEF , be two triangles having the two sides AB , AC ,

**equal**to the two sides DE , DF , each to each , viz . ... 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

The base DF is

The base DF is

**equal**to the base EF ; therefore ( I. 8. ) 3 . The angle DAF is**equal**to the angle EAF ; wherefore 4 . The angle BAC is bisected by the straight line**AF**. Which was to be done . PROP . X. - PROBLEM . Side 24

to B ; therefore the three straight lines KF , FG , GK , are

to B ; therefore the three straight lines KF , FG , GK , are

**equal**to the three A , B , C. And therefore 5 . ... DE , CE , so that CD be**equal**to**AF**, CE to AG , and DE to FG ; the angle FAG is**equal**to the angle DCE . Side 34

The alternate angles ABC , BCD , are

The alternate angles ABC , BCD , are

**equal**to one another ; and because AC is parallel to BD , and BC meets them , ( I. 29. ) ... and between the same parallels**AF**, BC ; the parallelogram ABCD shall be**equal**to the parallelogram EBCF .### 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...