Geometrical Problems Deducible from the First Six Books of Euclid, Arranged and Solved: To which is Added an Appendix Containing the Elements of Plane Trigonometry ...J. Smith, 1821 - 438 sider |
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Resultater 1-5 af 100
Side 10
... triangle ABC ; bisect the angles CAB , CBA by the lines AD , BD meeting in D , and draw DE , DF parallel to CA and CB respectively . AB will be trisected in E and F. A E F B Because ED is parallel to AC , the angle EDA = DAC = DAE and ...
... triangle ABC ; bisect the angles CAB , CBA by the lines AD , BD meeting in D , and draw DE , DF parallel to CA and CB respectively . AB will be trisected in E and F. A E F B Because ED is parallel to AC , the angle EDA = DAC = DAE and ...
Side 92
... triangle is greater than the differ- ence between the other two sides . Let ABC be a triangle ; any of its sides is greater than the difference of the other two . Let AC be greater than AB ; A B and cut off AD = AB ; join BD ; then the ...
... triangle is greater than the differ- ence between the other two sides . Let ABC be a triangle ; any of its sides is greater than the difference of the other two . Let AC be greater than AB ; A B and cut off AD = AB ; join BD ; then the ...
Side 93
... triangle perpendiculars be drawn to the sides ; they are together equal to a perpendicular drawn from any of the angles to the opposite side . From any point D within the ... triangle ABC ' be bisected Sect . 3. ] 93 GEOMETRICAL PROblems .
... triangle perpendiculars be drawn to the sides ; they are together equal to a perpendicular drawn from any of the angles to the opposite side . From any point D within the ... triangle ABC ' be bisected Sect . 3. ] 93 GEOMETRICAL PROblems .
Side 94
... triangle ABC ' be bisected in the points D , E , F ; join DE , EF , FD ; the triangle DEF is one fourth of the triangle ABC . A F B E Since AB and AC are bisected in D and F , ( Eucl . vi . 2. ) DF is parallel to BC ; and for the same ...
... triangle ABC ' be bisected in the points D , E , F ; join DE , EF , FD ; the triangle DEF is one fourth of the triangle ABC . A F B E Since AB and AC are bisected in D and F , ( Eucl . vi . 2. ) DF is parallel to BC ; and for the same ...
Side 95
... ABC be an isosceles triangle , having the side AB equal to AC . From B draw any line BD , and also BE cutting off DE equal to DB ; the angle ABD is double of CBE . A B C E For the angle DCB is equal to the two DEB , CBE , i . e . to the ...
... ABC be an isosceles triangle , having the side AB equal to AC . From B draw any line BD , and also BE cutting off DE equal to DB ; the angle ABD is double of CBE . A B C E For the angle DCB is equal to the two DEB , CBE , i . e . to the ...
Indhold
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49 | |
72 | |
108 | |
142 | |
187 | |
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293 | |
399 | |
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Almindelige termer og sætninger
ABCD angle ABC angle equal BC is equal centre chord circle cutting circumference construct the triangle cosecant cosine describe a circle divided equation equiangular Eucl extremities find the value given angle given circle given difference given in position given line given point given ratio given rectangle given square given straight line Hence inscribed intercepted isosceles triangle Join AE least common multiple Let AB Let ABC let fall line given line joining lines be drawn lines drawn mean proportional meeting opposite side parallel to AC parallelogram pendicular point of intersection produced quadrant radius rectangle contained right angles right-angled triangle segments semicircle shewn sine squares of AC tang tangent transposition trapezium triangle ABC triangle required vertex vertical angle whence
Populære passager
Side i - IF a straight line be divided into two equal, and also into two unequal parts ; the squares of the two unequal parts are together double of the square of half the line, and of the square of the line between the points of section.
Side xi - 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.
Side 319 - The circumference of every circle is supposed to be divided into 360 equal parts, called degrees ; each degree into 60 equal parts, called minutes ; and each minute into 60 equal parts, called seconds.
Side 150 - Iff a straight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the square of the other...
Side 204 - FC are equal to one another : wherefore the circle described from the centre F, at the distance of one of them, will pass through the extremities of the other two, and be described about the triangle ABC.
Side 115 - If from a point, without a parallelogram, there be drawn two straight lines to the extremities of the two opposite sides, between which, when produced, the point does not lie, the difference of the triangles thus formed is equal to half the parallelogram. Ex. 2. The two triangles, formed by drawing straight lines from any point within a parallelogram to the extremities of its opposite sides, are together half of the parallelogram.
Side 14 - In one of the given equations obtain the value of one of the unknown quantities in terms of the other unknown quantity; Substitute this value in the other equation and solve.
Side 291 - AB describe a segment of a circle containing an angle equal to the given angle, (in.
Side 297 - Given the vertical angle, the difference of the two sides containing it, and the difference of the segments of the base made by a perpendicular from the vertex ; construct the triangle.
Side 90 - If from any point in the base of an isosceles triangle perpendiculars...