## 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 5

Side 158

If a Perpendicular be drawn , in a Right - lined Triangle , from the Right Angle to

the Base , then the Triangles on each side of the Perpendicular are

to the whole , and also to one another . ET ABC be a Right - angled Triangle ...

If a Perpendicular be drawn , in a Right - lined Triangle , from the Right Angle to

the Base , then the Triangles on each side of the Perpendicular are

**similar**bothto the whole , and also to one another . ET ABC be a Right - angled Triangle ...

Side 171

fimilarly described Figure on the second ; which was to be demonstrated .

PROPOSITION XXI . THEOREM . Figures that are

lined Figure , are also

Figures A , B ...

fimilarly described Figure on the second ; which was to be demonstrated .

PROPOSITION XXI . THEOREM . Figures that are

**similar**to the same Right -lined Figure , are also

**similar**to one another . Le ET each of the Right - linedFigures A , B ...

Side 177

For the same Reason the Parallelogram ABCD is

. Therefore , both the Parallelograms EG , HK , are

ABCD . But Right - lined Figures that are

For the same Reason the Parallelogram ABCD is

**similar**to the Parallelogram KH. Therefore , both the Parallelograms EG , HK , are

**similar**to the ParallelogramABCD . But Right - lined Figures that are

**similar**to the same Right - lined Figure ... Side 180

To a Right Line given to apply a Parallelogram equal to a Right Line Figure given

, deficient by a Parallelogram , which is

but it is necessary that the Rightlined Figure given , to which the Parallelogram ...

To a Right Line given to apply a Parallelogram equal to a Right Line Figure given

, deficient by a Parallelogram , which is

**similar**to another given Parallelogram ;but it is necessary that the Rightlined Figure given , to which the Parallelogram ...

Side 181

Therefore XO is equal and

** 21 of thise

with FE : Let GPB be their Diameter , and the Figure be described . Then fince ...

Therefore XO is equal and

**similar**to KM , but KM is**similar**to EF , therefore XO is** 21 of thise

**similar**to EF , and so XO is about the fame Dia- † 26 of this . meterwith FE : Let GPB be their Diameter , and the Figure be described . Then fince ...

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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.