Euclid's Elements of Geometry: From the Latin Translation of Commandine. To which is Added, A Treatise of the Nature of Arithmetic of Logarithms ; Likewise Another of the Elements of Plain and Spherical Trigonometry ; with a Preface |
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Side 8
On the Same Right Line cannot be constituted two Right Lines equal to two other
Right Lines , each to each , at different Points , on the same Side , and having the
same Ends which the first Right Lines bave . FOB OR , if it be possible , let two ...
On the Same Right Line cannot be constituted two Right Lines equal to two other
Right Lines , each to each , at different Points , on the same Side , and having the
same Ends which the first Right Lines bave . FOB OR , if it be possible , let two ...
Side 35
Parallelograms constituted upon equal Bases , and between the same Parallels ,
are equal between themselves . L ET the Parallelograms ABCD , EFGH , be
between the fame Parallels AH , BG . I say , the Parallelogram ABCD is equal to
the ...
Parallelograms constituted upon equal Bases , and between the same Parallels ,
are equal between themselves . L ET the Parallelograms ABCD , EFGH , be
between the fame Parallels AH , BG . I say , the Parallelogram ABCD is equal to
the ...
Side 37
Equal Triangles constituted upon the same Base , on the Same Side , are in the
same Parallels . ÉT ABC , DB C , bè equal Triangles , constituted upon the fame
Base B C , on the fame Side . I say they are between the fame Parallels . For let ...
Equal Triangles constituted upon the same Base , on the Same Side , are in the
same Parallels . ÉT ABC , DB C , bè equal Triangles , constituted upon the fame
Base B C , on the fame Side . I say they are between the fame Parallels . For let ...
Side 38
And fo equal Triangles constituted upon equal Bases , on the same Side , are
between the same Parallels ' ; which was to be demonstrated . PROPOSITION
XLI . PROBLEM If a Parallelogram and a Triangle bave the same Base , and are
...
And fo equal Triangles constituted upon equal Bases , on the same Side , are
between the same Parallels ' ; which was to be demonstrated . PROPOSITION
XLI . PROBLEM If a Parallelogram and a Triangle bave the same Base , and are
...
Side 128
Equal Magnitudes have the same Proportion to tbe Tame Magnitude ; and one
and the same Magnitude has the same Proportion to equal Magnitudes . ET A , B
, be equal Magnitudes , and let C be any other Magnitude . I say , A and B have ...
Equal Magnitudes have the same Proportion to tbe Tame Magnitude ; and one
and the same Magnitude has the same Proportion to equal Magnitudes . ET A , B
, be equal Magnitudes , and let C be any other Magnitude . I say , A and B have ...
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Euclid's Elements of Geometry: From the Latin Translation of Commandine. To ... John Keill Uten tilgangsbegrensning - 1723 |
Vanlige uttrykk og setninger
added alſo Altitude Angle ABC Baſe becauſe Center Circle Circle ABCD Circumference common Cone conſequently contained Cylinder demonſtrated deſcribed Diameter Difference Diſtance divided double draw drawn equal equal Angles equiangular Equimultiples exceeds fall fame firſt fore four fourth given greater half join leſs likewiſe Logarithm Magnitudes Manner mean Multiple Number oppoſite parallel Parallelogram perpendicular Place Plane Point Polygon Priſms produced Prop Proportion PROPOSITION proved Pyramid Radius Ratio Rectangle remaining Right Angles Right Line Right-lined Figure ſaid ſame ſame Reaſon ſay ſecond Segment Series ſhall ſhall be equal Sides ſimilar ſince Sine Solid ſome Sphere Square ſtand taken Terms THEOREM thereof theſe third thoſe thro touch Triangle Triangle ABC Unity Wherefore whole whoſe Baſe
Populære avsnitt
Side 66 - 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 161 - 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 110 - And in like manner it may be shown that each of the angles KHG, HGM, GML is equal to the angle HKL or KLM ; therefore the five angles GHK, HKL, KLM, LMG, MGH...
Side 88 - IN a circle, the angle in a semicircle is a right angle ; but the angle in a segment greater than a semicircle is less than a right angle ; and the angle in a segment less than a semicircle is greater than a right angle.
Side 22 - ... 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...
Side 9 - ... 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 equal to DE, and AC to DF ; but the base CB greater than the base EF ; the angle BAC is likewise greater than the angle EDF.
Side 15 - CF, and the triangle AEB to the triangle CEF, and the remaining angles to the remaining angles, each to each, to which...
Side 33 - ... therefore their other sides are equal, each to each, and the third angle of the one to the third angle of the other, (i.
Side 111 - 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.