## Euclid's Elements of Geometry: From the Latin Translation of Commandine. To which is Added, A Treatise of the Nature of Arithmetic of Logarithms ; Likewise Another of the Elements of Plain and Spherical Trigonometry ; with a Preface |

### Inni boken

Resultat 1-5 av 6

Side 56

Therefore the

is equal to GF , and the

Squares of EG , GF , together , are double to the

Therefore the

**Square**of E A is double to the**Square**of AC . Again , because EGis equal to GF , and the

**Square**of E G is equal to the**Square**of GF : Therefore theSquares of EG , GF , together , are double to the

**Square**of GF . But the**Square**... Side 58

Ia But the

Therefore the Squares of AE , EG , are double the Squares of AC , CD . But the

of AG ...

Ia But the

**Square**of E A has been proved to be double to the**Square**of A C.Therefore the Squares of AE , EG , are double the Squares of AC , CD . But the

**Square**of AG ist equal to the Squares of AE , EG ; and consequently the**Square**of AG ...

Side 97

with the

to EB ; wherefore the Rectangle under AD , DC , together with the

is equal to the

with the

**Square**of EC , shall * be equal the**Square*** 6. 20 of ED . But EC is equalto EB ; wherefore the Rectangle under AD , DC , together with the

**Square**of EB ,is equal to the

**Square**of ED . But the**Square**of ED is – equal to the**Square**of ... Side 241

Wherefore as the

BOCPDR to * the Polygon EKFLGMHN . But * 1 of tbiso as the

the

...

Wherefore as the

**Square**of B D is to the**Square**of FH , so is the Polygon AXBOCPDR to * the Polygon EKFLGMHN . But * 1 of tbiso as the

**Square**of BD is tothe

**Square**of FH , so is the Circle ABCD to the Space S. Wherefore as the Circle...

Side 271

And because AB is equal to AK , the

the Angle att 47. 1 . Z is a Right Angle . And the Squares of A Z , ZK , are equal to

the ...

And because AB is equal to AK , the

**Square**of AB shall be also equal to the**Square**of AK : And the Squares of AZ , ŽB are f equal to the**Square**of AB . Forthe Angle att 47. 1 . Z is a Right Angle . And the Squares of A Z , ZK , are equal to

the ...

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Euclid's Elements of Geometry: From the Latin Translation of Commandine. To ... John Keill Uten tilgangsbegrensning - 1723 |

### Vanlige uttrykk og setninger

added alſo Altitude Angle ABC Baſe becauſe Center Circle Circle ABCD Circumference common Cone conſequently contained Cylinder demonſtrated deſcribed Diameter Difference Diſtance divided double draw drawn equal equal Angles equiangular Equimultiples exceeds fall fame firſt fore four fourth given greater half join leſs likewiſe Logarithm Magnitudes Manner mean Multiple Number oppoſite parallel Parallelogram perpendicular Place Plane Point Polygon Priſms produced Prop Proportion PROPOSITION proved Pyramid Radius Ratio Rectangle remaining Right Angles Right Line Right-lined Figure ſaid ſame ſame Reaſon ſay ſecond Segment Series ſhall ſhall be equal Sides ſimilar ſince Sine Solid ſome Sphere Square ſtand taken Terms THEOREM thereof theſe third thoſe thro touch Triangle Triangle ABC Unity Wherefore whole whoſe Baſe

### Populære avsnitt

Side 66 - 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.

Side 161 - IF two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals : the triangles shall be equiangular, and shall have those angles equal which are opposite to the homologous sides.

Side 110 - And in like manner it may be shown that each of the angles KHG, HGM, GML is equal to the angle HKL or KLM ; therefore the five angles GHK, HKL, KLM, LMG, MGH...

Side 88 - IN a circle, the angle in a semicircle is a right angle ; but 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 22 - ... sides equal to them of the other. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB...

Side 9 - ... equal to them, of the other. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB equal to DE, and AC to DF ; but the base CB greater than the base EF ; the angle BAC is likewise greater than the angle EDF.

Side 15 - CF, and the triangle AEB to the triangle CEF, and the remaining angles to the remaining angles, each to each, to which...

Side 33 - ... therefore their other sides are equal, each to each, and the third angle of the one to the third angle of the other, (i.

Side 111 - If two right-angled triangles have their hypotenuses equal, and one side of the one equal to one side of the other, the triangles are congruent.