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, 1827 - 377 σελίδες |
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Σελίδα 2
... Join AD , DB . Since AC = CB and CD is common , and the angle ACD = BCD being right angles , AD A E = DB . And the same may be proved of lines drawn from any other point in CD to A and B. But if a point E be taken which is not in CD , join ...
... Join AD , DB . Since AC = CB and CD is common , and the angle ACD = BCD being right angles , AD A E = DB . And the same may be proved of lines drawn from any other point in CD to A and B. But if a point E be taken which is not in CD , join ...
Σελίδα 3
... Join AE , EB . Since AF FB , and FE is common , and the angles at Fare right angles , therefore AE = EB . A E F F B ( 5. ) From two given points on the same side of a line given in position , to draw two lines which shall meet in that ...
... Join AE , EB . Since AF FB , and FE is common , and the angles at Fare right angles , therefore AE = EB . A E F F B ( 5. ) From two given points on the same side of a line given in position , to draw two lines which shall meet in that ...
Σελίδα 5
... join AE . Then AE = AF ( i . 2. ) . And besides AE no other line can be drawn equal to AF . For , if possible , let AI = AF . Then be- cause AI - AF and AF AE , therefore AI = AE , 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 AI = AF . Then be- cause AI - AF and AF AE , therefore AI = AE , i . e . a line more remote is equal to one nearer the ...
Σελίδα 7
... AE be drawn equal to AC , inclined at any angle to AB ; join EB , EC , ED ; the angle BEC B is equal to the angle CED . E D For since AB : AC :: AC : AD , and AE = AC ; : . 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 B is equal to the angle CED . E D For since AB : AC :: AC : AD , and AE = AC ; : . AB : AE :: AE : AD , i.e. the sides about the angle A are ...
Σελίδα 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 ...
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Άλλες εκδόσεις - Προβολή όλων
Geometrical Problems Deducible from the First Six Books of Euclid: Arranged ... Miles Bland Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2015 |
Geometrical Problems Deducible From the First Six Books of Euclid: Arranged ... Miles Bland Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2023 |
Geometrical Problems Deducible from the First Six Books of Euclid: Arranged ... Miles Bland Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2018 |
Συχνά εμφανιζόμενοι όροι και φράσεις
ABCD angle ABC base bisect the angle centre chord circle ABC circles cut circumference describe a circle divided draw a line 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 intercepted isosceles triangle Join AB Join AE Join BD Let AB Let ABC let fall line given line joining line required lines be drawn lines drawn mean proportional opposite sides parallel to AC parallelogram pendicular point of bisection point of contact point of intersection radius rectangle rectangle contained right angles right-angled triangle segments semicircle shewn tangent touching the circle trapezium triangle ABC whence
Δημοφιλή αποσπάσματα
Σελίδα 10 - 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.
Σελίδα xv - 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.
Σελίδα xxx - AB be the given straight line ; it is required to divide it into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part.
Σελίδα 303 - 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.
Σελίδα 140 - 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...
Σελίδα 329 - CE is equal to the difference of the segments of the base made by the perpendicular.
Σελίδα 109 - 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.
Σελίδα 164 - PROPOSITION I. PROBLEM. — To describe an equilateral triangle upon a given finite straight line. Let AB be the given straight line; it is required to describe an equilateral triangle upon it.
Σελίδα 281 - 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.
Σελίδα 270 - AB describe a segment of a circle containing an angle equal to the given angle, (in.