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 100
Side 2
... drawn from a certain Point within the Figure , are equal . XVI . And that Point is called the Center of the Circle . XVII . A Diameter of a Circle , is a Right Line drawn through the Center , and terminated on both Sides by the ...
... drawn from a certain Point within the Figure , are equal . XVI . And that Point is called the Center of the Circle . XVII . A Diameter of a Circle , is a Right Line drawn through the Center , and terminated on both Sides by the ...
Side 5
... draw the Right Lines CA , CBt . Then because A is the Center of the Circle DBC , AC fhall be equal to AB . And ... Draw + Poft . 15 Def . Poft . 1. Draw the Right Line AC from the Book I. Euclid's ELEMENTS . 5.
... draw the Right Lines CA , CBt . Then because A is the Center of the Circle DBC , AC fhall be equal to AB . And ... Draw + Poft . 15 Def . Poft . 1. Draw the Right Line AC from the Book I. Euclid's ELEMENTS . 5.
Side 13
... draw a Right Line from the Point C , at Right Angles to A B. Affume any Point D in AC , and make C E equal * to CD , and upon DE make † the Equilateral Tri- * 3 of this . angle FDE , and join FC . I fay , the Right Linet FC is drawn ...
... draw a Right Line from the Point C , at Right Angles to A B. Affume any Point D in AC , and make C E equal * to CD , and upon DE make † the Equilateral Tri- * 3 of this . angle FDE , and join FC . I fay , the Right Linet FC is drawn ...
Side 14
... drawn perpendicular , upon a given infinite Right Line , from a given Point out of it ; which was to be demonstrated ... draw BE from the Point B , at Right Angles to CD . Therefore the Angles CBE , EBD , are two Right Angles : And ...
... drawn perpendicular , upon a given infinite Right Line , from a given Point out of it ; which was to be demonstrated ... draw BE from the Point B , at Right Angles to CD . Therefore the Angles CBE , EBD , are two Right Angles : And ...
Side 15
... drawn from contrary Parts , making the ad- jacent Angles , both together , equal to two Right An- gles , the faid two Right Lines will make but one Straight Line . FOR let two Right Lines BC , BD , drawn from contrary Parts to the Point ...
... drawn from contrary Parts , making the ad- jacent Angles , both together , equal to two Right An- gles , the faid two Right Lines will make but one Straight Line . FOR let two Right Lines BC , BD , drawn from contrary Parts to the Point ...
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.