## The Elements of Euclid |

### Inni boken

Resultat 1-5 av 12

Side 85

to AC ; for the same reason , AC is parallel to FK , and in like

may each of them be demonstrated to be parallel to BED ; therefore the figures

GK , GC , AK , FB , BK are parallelograms ; and GF is therefore equal ( 34 . 1 . ) to

HK ...

to AC ; for the same reason , AC is parallel to FK , and in like

**manner**GF , HKmay each of them be demonstrated to be parallel to BED ; therefore the figures

GK , GC , AK , FB , BK are parallelograms ; and GF is therefore equal ( 34 . 1 . ) to

HK ...

Side 89

angle FLC : and because KC is equal to CL , KL is double of KC : in the same

KC , as was demonstrated , and that KL is double of KC , and HK double of BK ,

HK ...

angle FLC : and because KC is equal to CL , KL is double of KC : in the same

**manner**, it may be shown that HK is double of BK : and because BK is equal toKC , as was demonstrated , and that KL is double of KC , and HK double of BK ,

HK ...

Side 113

In like

remainder LN is the same multiple of the remainder CF , that the whole LM is of

the KL whole CD ( 5 . 5 . ) : but it was shown that LM is the same multiple of CD ,

that ...

In like

**manner**, because LM is the same multiple of CD , that MN is of DF , theremainder LN is the same multiple of the remainder CF , that the whole LM is of

the KL whole CD ( 5 . 5 . ) : but it was shown that LM is the same multiple of CD ,

that ...

Side 177

O , P . In the same

straight line DH with NR , and the point H with the point R : and because the solid

angle at B is equal to the solid angle at L , it may be proved , in the same

...

O , P . In the same

**manner**, the figure AH coincides with the figure KR , and thestraight line DH with NR , and the point H with the point R : and because the solid

angle at B is equal to the solid angle at L , it may be proved , in the same

**manner**...

Side 193

In the same

is equal to DF ; therefore , since AB is equal to DE , BA and AC are equal to ED

and DF ; and the angle BAC is equal to the angle EDF ; wherefore the base BC is

...

In the same

**manner**, if HC and MF be joined , it may be demonstrated that AC Eis equal to DF ; therefore , since AB is equal to DE , BA and AC are equal to ED

and DF ; and the angle BAC is equal to the angle EDF ; wherefore the base BC is

...

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### Vanlige uttrykk og setninger

added altitude angle ABC angle BAC base BC is given centre circle circle ABCD circumference common cone contained cylinder definition demonstrated described diameter divided double draw drawn equal equal angles equiangular equimultiples Euclid excess fore four fourth given angle given in magnitude given in position given in species given magnitude given ratio given straight line gles greater Greek half join less likewise magnitude manner meet multiple Note opposite parallel parallelogram pass perpendicular plane prism produced PROP proportionals proposition pyramid Q. E. D. PROP reason rectangle rectangle contained remaining right angles segment shown sides similar sine solid sphere square square of BC taken THEOR third triangle ABC wherefore whole

### Populære avsnitt

Side 36 - 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. Let the straight line AB be divided into any two parts at the point C : the squares of AB, BC shall be equal to twice the rectangle AB, BC, together with the square of AC.

Side 145 - Wherefore, in equal circles &c. QED PROPOSITION B. THEOREM If the vertical angle of a triangle be bisected by a straight line which likewise cuts the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square on the straight line which bisects the angle.

Side 65 - The angle in a semicircle is a right angle; the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle.

Side 248 - Again ; the mathematical postulate, that " things which are equal to the same are equal to one another," is similar to the form of the syllogism in logic, which unites things agreeing in the middle term.

Side 11 - If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles.

Side 121 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.

Side 21 - 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 14 - To construct a triangle of which the sides shall be equal to three given straight lines ; but any two whatever of these lines must be greater than the third (20.

Side 80 - 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 133 - ... rectilineal figures are to one another in the duplicate ratio of their homologous sides.