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, 1819 - 377 sider |
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Side 3
... Join AE , EB . Since AF = FB , and FE is common , and the angles at F are right angles , therefore AE = EB . F B F E D B ( 5. ) From two given points on the same side of a line given in position , to draw two lines which shall meet in ...
... Join AE , EB . Since AF = FB , and FE is common , and the angles at F are right angles , therefore AE = EB . F B F E D B ( 5. ) From two given points on the same side of a line given in position , to draw two lines which shall meet in ...
Side 5
... join AE . Then AE = AF ( i . 2. ) . And besides AE no other line can be drawn equal to AF . For , if possible , let AIAF . Then be- cause AI - AF and AF - AE , therefore AIAE , i . e . a line more remote is equal to one nearer the ...
... join AE . Then AE = AF ( i . 2. ) . And besides AE no other line can be drawn equal to AF . For , if possible , let AIAF . Then be- cause AI - AF and AF - AE , therefore AIAE , i . e . a line more remote is equal to one nearer the ...
Side 9
... AE be drawn equal to AC , inclined at any angle to AB ; join EB , EC , ED ; the angle BEC is equal to the angle CED . E D For since AB : AC :: AC : AD , and AE = AC ; 1 : . AB : AE :: AE : AD , i . e . the sides about the angle A are ...
... AE be drawn equal to AC , inclined at any angle to AB ; join EB , EC , ED ; the angle BEC is equal to the angle CED . E D For since AB : AC :: AC : AD , and AE = AC ; 1 : . AB : AE :: AE : AD , i . e . the sides about the angle A are ...
Side 20
... Join AF ; and draw BK parallel to AG cutting AF in L ; and draw LM parallel to KE cutting AE in M and AG in N. Then FE : LM :: GF : ( NL = ) AB and FE DG :: FG : AB by construction ; : .. LM = DG = IA , if therefore ILO be drawn , IL ...
... Join AF ; and draw BK parallel to AG cutting AF in L ; and draw LM parallel to KE cutting AE in M and AG in N. Then FE : LM :: GF : ( NL = ) AB and FE DG :: FG : AB by construction ; : .. LM = DG = IA , if therefore ILO be drawn , IL ...
Side 21
... Join AE , FE , and HC parallel Then ( Eucl . vi . 2. ) AI IH :: AB : BE in the given ratio of the remainders ; and ... AE to meet CB in E making the angle AEC = the given angle to be made by the line to be drawn , with BC . In AE take Ad ...
... Join AE , FE , and HC parallel Then ( Eucl . vi . 2. ) AI IH :: AB : BE in the given ratio of the remainders ; and ... AE to meet CB in E making the angle AEC = the given angle to be made by the line to be drawn , with BC . In AE take Ad ...
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Andre utgaver - Vis alle
Geometrical Problems Deducible from the First Six Books of Euclid, Arranged ... Miles Bland Uten tilgangsbegrensning - 1821 |
Geometrical Problems Deducible from the First Six Books of Euclid, Arranged ... Miles Bland Uten tilgangsbegrensning - 1819 |
Geometrical problems deducible from the first six books of Euclid, arranged ... Miles Bland Uten tilgangsbegrensning - 1819 |
Vanlige uttrykk og setninger
ABCD angle ABC angles at F base centre chord circle ABC circles cut circles touch circumference describe a circle divided draw a line draw the diameter drawn parallel duplicate ratio equal angles equiangular Eucl extremities G draw given angle given circle given in position given line given point given ratio given square given straight line given triangle intercepted isosceles triangle Join AE Join BD Let AB Let ABC let fall line given line joining line required lines be drawn lines drawn mean proportional opposite side parallel to BC parallelogram pendicular point of bisection point of contact point of intersection point required radius rectangle right angles segments semicircle shewn tangent touching the circle trapezium triangle ABC
Populære avsnitt
Side 14 - 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 xiii - 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 230 - To describe an isosceles triangle, having each of the angles at the base double of the third angle.
Side 327 - 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 158 - 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 212 - 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 123 - 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 305 - 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 247 - The perpendicular from the vertex on the base of an equilateral triangle is equal to the side of an equilateral triangle inscribed in a circle, whose diameter is the base.
Side 299 - AB be equal to the given bisecting line ; and upon it describe a segment of a circle containing an angle equal to the given angle.