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 páginas |
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... Reason , that I am eafily in- duced to believe the Obfcurity , Sciolifts fo often accufed Euclid with , is rather to be attributed to their own perplex'd Ideas , than to the Demonftrations themselves . And however Some may find Fault ...
... Reason , that I am eafily in- duced to believe the Obfcurity , Sciolifts fo often accufed Euclid with , is rather to be attributed to their own perplex'd Ideas , than to the Demonftrations themselves . And however Some may find Fault ...
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... Reason I know not , have entirely omitted fome Propofitions , and have al- tered the Demonftrations of others for worfe . Among whom are chiefly Tacquet and Dechalles , both of which have un . happily rejected fome elegant Propofitions ...
... Reason I know not , have entirely omitted fome Propofitions , and have al- tered the Demonftrations of others for worfe . Among whom are chiefly Tacquet and Dechalles , both of which have un . happily rejected fome elegant Propofitions ...
Página 35
... Reason , the Pa- rallelogram 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 ...
... Reason , the Pa- rallelogram 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 ...
Página 69
... Reason , CF is greater than F G. For the fame + 24. 1 . Again , becaufe G F and F E are † greater than GE , and G E is equal to ED ; GF and FE fhall be greater than ED ; and if FE , which is com- mon , be taken away , then the Remainder ...
... Reason , CF is greater than F G. For the fame + 24. 1 . Again , becaufe G F and F E are † greater than GE , and G E is equal to ED ; GF and FE fhall be greater than ED ; and if FE , which is com- mon , be taken away , then the Remainder ...
Página 74
... Reason , the Center of the Circle will be in OL . And the Right Lines HK , OL , have no other Point common but D : Therefore D is the Center of the Circle ABC ; which was to be demonftrated . PROPOSITION X. THEOREM . A Circle cannot cut ...
... Reason , the Center of the Circle will be in OL . And the Right Lines HK , OL , have no other Point common but D : Therefore D is the Center of the Circle ABC ; which was to be demonftrated . PROPOSITION X. THEOREM . A Circle cannot cut ...
Términos y frases comunes
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
Pasajes populares
Página 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.
Página 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.
Página 63 - DBA ; and because AE, a side of the triangle DAE, is produced to B, the angle DEB is greater (16.
Página 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...
Página 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.
Página 17 - CF, and the triangle AEB to the triangle CEF, and the remaining angles to the remaining angles, each to each, to which...
Página 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...
Página 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.
Página 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.
Página 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.