## 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 10

Side 111

Then

the Right Line FC is drawn from the Center F to C , the Point of ConFC will be I

perpendicular to KL : And so I 18. 36 both the Angles at Care Right Angles . For

the ...

Then

**because**the Right Line K L touches the Circle ABCDE in the Point C , andthe Right Line FC is drawn from the Center F to C , the Point of ConFC will be I

perpendicular to KL : And so I 18. 36 both the Angles at Care Right Angles . For

the ...

Side 153

Set the Side B.C , in the fame Right Line with the Side CE , and

Angles ABC , ACB , are * less than two Right Angles , and the Angle ACB * 19. 1 .

is equal to the Angle DEC , the Angles ABC , DEC , are less than two Right

Angles ...

Set the Side B.C , in the fame Right Line with the Side CE , and

**because**theAngles ABC , ACB , are * less than two Right Angles , and the Angle ACB * 19. 1 .

is equal to the Angle DEC , the Angles ABC , DEC , are less than two Right

Angles ...

Side 209

Then ,

to each and they contain equal Angles , the Base AL Ihall be equal to the Base

GK . And since the Angles E and H are greater than the Angle ABC , whereof the

...

Then ,

**because**the two Sides AB , BL , are equal to the two Sides GH , HK , eachto each and they contain equal Angles , the Base AL Ihall be equal to the Base

GK . And since the Angles E and H are greater than the Angle ABC , whereof the

...

Side 217

Then

shall be * perpendicular to all the Right ... For the same Reason , both the Angles

HKA , HKB , are Right Angles ; and

the ...

Then

**because**F G is perpendicular to the Plane passing through ED , DC , itshall be * perpendicular to all the Right ... For the same Reason , both the Angles

HKA , HKB , are Right Angles ; and

**because**the two Sides KA , AB , are equal tothe ...

Side 270

And

parallel to KB : But BO , KS , join them . Therefore KBOS is * a quadrilateral

Figure in one Plane : For if two Right Lines be parallel , and Points be taken in

both of ...

And

**because**QV is parallel to SO , and also parallel to KB , SO shall be also Iparallel to KB : But BO , KS , join them . Therefore KBOS is * a quadrilateral

Figure in one Plane : For if two Right Lines be parallel , and Points be taken in

both of ...

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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.