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 ...T. Woodward, 1723 - 364 sider |
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Resultat 1-5 av 15
Side 35
... EFGH is equal to the fame Parallelo- gram EBCH . Therefore the Parallelogram ABCD fhall be equal to the Parallelogram EFGH . And fo Parallelograms conftituted upon equal Bafes , and be- tween the fame Parallels , are equal between ...
... EFGH is equal to the fame Parallelo- gram EBCH . Therefore the Parallelogram ABCD fhall be equal to the Parallelogram EFGH . And fo Parallelograms conftituted upon equal Bafes , and be- tween the fame Parallels , are equal between ...
Side 239
... EFGH , be Circles , whofe Dia- meters are BD , FH . I fay , as the Square of BD is to the Square of FH , fo the Circle ABCD to the Circle EFGH . For * 41. 1 . For if it be not fo Book XII . Euclid's ELEMENTS 239.
... EFGH , be Circles , whofe Dia- meters are BD , FH . I fay , as the Square of BD is to the Square of FH , fo the Circle ABCD to the Circle EFGH . For * 41. 1 . For if it be not fo Book XII . Euclid's ELEMENTS 239.
Side 240
... EFGH . First let it be to a Space S , lefs than the Circle E FGH , and let the Square EFGH be defcribed therein . * This Square EFGH will be greater than half the Circle EFGH ; because if we draw Tangents to the Circle thro ' the Points ...
... EFGH . First let it be to a Space S , lefs than the Circle E FGH , and let the Square EFGH be defcribed therein . * This Square EFGH will be greater than half the Circle EFGH ; because if we draw Tangents to the Circle thro ' the Points ...
Side 241
... EFGH . After the fame Manner we likewife demonftrate that the Square of FH to the Square of BD is not as the Cir- cle EFGH , to fome Space lefs than the Circle AB- CD . Laftly , I fay , the Square of BD to the Square of FH is not as the ...
... EFGH . After the fame Manner we likewife demonftrate that the Square of FH to the Square of BD is not as the Cir- cle EFGH , to fome Space lefs than the Circle AB- CD . Laftly , I fay , the Square of BD to the Square of FH is not as the ...
Side 258
... EFGH , fo is the Cone A L to fome Solid either lefs or greater than the Cone EN . First , let it be to the Solid X ... EFGH be defcribed in the Circle EFGH , which Square is greater than one half of the Circle , and erect a Pyramid upon ...
... EFGH , fo is the Cone A L to fome Solid either lefs or greater than the Cone EN . First , let it be to the Solid X ... EFGH be defcribed in the Circle EFGH , which Square is greater than one half of the Circle , and erect a Pyramid upon ...
Vanlige uttrykk og setninger
alfo equal alſo Angle ABC Angle BAC Baſe becauſe bifected Center Circle ABCD Circle EFGH Circumference Cofine Cone confequently contain'd Coroll Cylinder defcrib'd defcribed demonftrated Diameter Diſtance drawn thro equal Angles equiangular equilateral Equimultiples faid fame Altitude fame Multiple fame Plane fame Proportion fame Reafon fecond fhall be equal fimilar fince firft firſt folid Parallelepipedon fome fore ftand fubtending given Right Line Gnomon greater join leffer lefs likewife Logarithm Magnitudes Meaſure Number paffing thro Parallelogram perpendicular Polygon Prifm Priſms Prop PROPOSITION Pyramid Quadrant Ratio Rectangle remaining Angle Right Angles Right Line A B Right Line AB Right-lin'd Figure Right-lin❜d Segment ſhall Sine Solid Sphere Subtangent thefe THEOREM theſe thofe Triangle ABC triplicate Proportion Unity Vertex the Point Wherefore whofe Bafe whole
Populære avsnitt
Side 190 - 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 160 - 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 63 - DBA ; and because AE, a side of the triangle DAE, is produced to B, the angle DEB is greater (16.
Side 152 - ... 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...
Side 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.
Side 17 - CF, and the triangle AEB to the triangle CEF, and the remaining angles to the remaining angles, each to each, to which...
Side 210 - 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...
Side 229 - 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.
Side 164 - 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.
Side 93 - 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.