## 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 Plain and Spherical Trigonometry : with a Preface ... |

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Page 340

... preffes the Length of the

... preffes the Length of the

**Subtangent**AT . Fig . 4 . After Neper's Death , the excellent Mr. Henry Briggs , by great Pains , made and published Tables of Loga- rithms according to this Form . Now fince in these rithms 340 OF LOGARITHMS . Page 356

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**Subtangent**, will every where be a conftant and given Length . And this is one of the chief Proper- ties of the logarithmical Curve ; for the different Species or Forms of those Curves are determined by the Subtangents . م called The ... Page 357

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**Subtangent**AT is to the**Subtangent**MX . For let there be supposed an infinite Number of mean proportional Terms between AB , CD , or NM , QY , in the Ratio of AB to ab , or MN to mn ; and fince AB = MN , then will ab = mn , as also bono ... Page 358

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**Subtangent**thereof MX shall be conftant and invariable , and the small Triangle Non shall be similar to the Triangle XMN . But it has been prov'd , that the Increment no = Mm xa = Noxa ; and sono : No :: Noxa : No :: a : 1. But as no is ... Page 359

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**Subtangent**of the Lo- garithmical Curve expressing Brigg's Logarithms ) so is one Year or Unity to a Briggian Logarithm , answering to the Number QY . This Logarithm will be found 0. 0217147 , and the Number answering to the same is ...### Common terms and phrases

alfo alſo equal Altitude Angle ABC Angle BAC Angle BCD Bafe Baſe BC becauſe biſected Center Circle ABCD Circle EFGH Circumference Cofine Cone confequently contain'd Coroll Cylinder defcrib'd DEFH demonftrated deſcribed Diameter Diſtance drawn thro equal Angles equiangular equilateral Equimultiples fame Multiple fame Reaſon fimilar fince firſt folid Parallelepipedon fore fubtending given Right Line greater join leffer leſs likewife Logarithm Magnitudes Meaſure Number oppofite parallel Parallelogram perpendicular Plane Polygon Priſms Prop PROPOSITION Pyramid Quadrant Ratio Rectangle Rectangle contained remaining Angle Right Angles Right Line AC Right-lin'd Figure ſame ſay ſecond Segment Semicircle ſhall be equal Sine ſome Sphere ſtand Subtangent THEOREM theſe thoſe Triangle ABC Unity Vertex the Point Wherefore whole

### Popular passages

Page 192 - 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 162 - 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 63 - DBA ; and because AE, a side of the triangle DAE, is produced to B, the angle DEB is greater (16.

Page 154 - ... therefore the angle DFG is equal to the angle DFE, and the angle at G to the angle at E : but the angle DFG is equal to the angle ACB...

Page 100 - About a given circle to describe a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given triangle; it is required to describe a triangle about the circle ABC equiangular to the triangle DEF.

Page 17 - CF, and the triangle AEB to the triangle CEF, and the remaining angles to the remaining angles, each to each, to which...

Page 212 - 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 233 - 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.

Page 166 - ABG ; (vi. 1.) therefore the triangle ABC has to the triangle ABG the duplicate ratio of that which BC has to EF: but the triangle ABG is equal to the triangle DEF; therefore also the triangle ABC has to the triangle DEF the duplicate ratio of that which BC has to EF. Therefore similar triangles, &c.

Page 95 - If from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it ; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square of the line which touches it.