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

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Page 169

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**Polygons**are divided into fimilar Trian- gles , equal in Number , and homologous to the Wholes ; and**Polygon**to**Polygon**, is in the du- plicate Proportion of that which one homolo- gous Side has to the other . LET ABCDE , FGHKL , be ... Page 170

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**Polygon**ABCDE , to the**Polygon**F G H KL , is in the du plicate Proportion of an homologous Side of the one , to an homologous Side of the other ; that is , as A B to FG . For , because the Triangle A BE is fimilar to the 19 of this ... Page 171

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**Polygon**ABCDE , to the**Polygon**F G H K L , is in the du- plicate Proportion of the homologous Side A B to the homologous Side F G. Therefore , fimilar**Polygons**are divided into fimilar Triangles , equal in Number , and ho- mologous to ... Page 238

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**Polygon**ABCDE to the**Polygon**FGHKL . For , join BE , AM , GL , FN . Then , because the**Polygon**ABCDE is fimilar to the**Polygon**FGHKL , the Angle BAE is equal to the Angle GFL ; and BA is to AE , as GF is to FL : Therefore the two ... Page 239

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**Polygon**ABCDE to the**Polygon**FGHK L. Therefore , fimilar**Polygons**, infcribed in Circles , are to one another as the Squares of the Diameters of the Circles ; which was to be demonftrated . LEMM A. D A A F + 1 + K H + If there be two ...### Other editions - View all

Euclid's Elements of Geometry: From the Latin Translation of Commandine. to ... John Keill No preview available - 2018 |

Euclid's Elements of Geometry: From the Latin Translation of Commandine. to ... John Keill No preview available - 2017 |

Euclid's Elements of Geometry: From the Latin Translation of Commandine. to ... John Keill No preview available - 2015 |

### Common terms and phrases

A B C adjacent Angles alfo equal alſo Angle ABC Baſe becauſe bifected Centre Circle ABCD Circumference Cofine Cone confequently Coroll Cylinder defcribed demonftrated Diameter Diſtance drawn thro equal Angles equiangular Equimultiples faid fame Altitude fame Multiple fame Plane fame Proportion fame Reaſon fecond fhall be equal fimilar fince firft folid Parallelepipedon fome fore ftand fubtending given Right Line Gnomon greater join leffer lefs likewife Logarithm Magnitudes Meaſure Number oppofite parallel Parallelogram perpendicular Polygon Prifm Prop PROPOSITION Pyramid Quadrant Ratio Reafon Rectangle Rectangle contained remaining Angle Right Angles Right Line A B Right-lined Figure Segment ſhall Sides A B Sine Square Subtangent thefe THEOREM thofe thoſe tiple Triangle ABC Unity Vertex the Point Wherefore whofe Bafe whoſe

### Popular passages

Page 195 - 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.

Page 165 - 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.

Page 169 - Equal parallelograms which have one angle of the one equal to one angle of the other, have their sides about the equal angles...

Page xxii - ... 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...

Page 54 - If a straight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the square of the other part, is equal to the square of the straight line which is made up of the whole and that part.

Page 123 - GB is equal to E, and HD to F; GB and HD together are equal to E and F together : wherefore as many magnitudes as...

Page 215 - CD; therefore AC is a parallelogram. In like manner, it may be proved that each of the figures CE, FG, GB, BF, AE, is a parallelogram...

Page 196 - ABC, and they are both in the same plane, which is impossible ; therefore the straight line BC is not above the plane in which are BD and BE: wherefore, the three straight lines BC, BD, BE are in one and the same plane. Therefore, if three straight lines, &c.

Page 161 - And because HE is parallel to KC, one of the sides of the triangle DKC, as CE to ED, so is...

Page 207 - A plane is perpendicular to a plane, when the straight lines drawn in one of the planes perpendicular to the common section of the two planes, are perpendicular to the other plane. V. The inclination of a straight line to a plane...