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 |
Fra bogen
Resultater 1-5 af 100
Side 3
... Join AE , EB . Since AF FB , and FE is common , and the angles at F are right angles , therefore AE = EB . A F B F E Ꭰ 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 . A F B F E Ꭰ 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
... AE = = And from A there can only be drawn to BC two straight lines equal to each other , viz . one of AD . Make DE = DF and join AE . AF ( i . 2. ) . And besides AE no other line can be drawn equal to AF . For , if possible , let AI ...
... AE = = And from A there can only be drawn to BC two straight lines equal to each other , viz . one of AD . Make DE = DF and join AE . AF ( i . 2. ) . And besides AE no other line can be drawn equal to AF . For , if possible , let AI ...
Side 9
... 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 C D = AC ; For since AB : AC :: AC : AD , and AE = . .. AB : AE :: AE : AD , i . e . the sides about the angle A ...
... 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 C D = AC ; For since AB : AC :: AC : AD , and AE = . .. AB : AE :: AE : AD , i . e . the sides about the angle A ...
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 ...
... 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 ...
Indhold
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Almindelige termer og sætninger
ABCD angle ABC base bisects the angle centre chord circle ABC circles cut circles touch circumference describe a circle divided draw any line drawn parallel duplicate ratio equal angles equiangular Eucl extremities given angle given circle given in position given line given point given ratio given straight line given triangle inscribed 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 perpendicular be drawn point of bisection point of contact point of intersection quadrant radius rectangle contained right angles right-angled triangle segments semicircle shewn tangent touches the circle trapezium triangle ABC vertex vertical angle
Populære passager
Side 124 - If a straight line be drawn parallel to one of the sides of a triangle, it shall cut the other sides, or those sides produced, proportionally; and if the sides, or the sides produced, be cut proportionally, the straight '.line which joint the points of section, shall be parallel to the remaining side of the triangle.
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 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 160 - Upon a given straight line, to describe a segment of a circle, containing an angle equal to a given angle. Let AB be the given straight line, and C the given angle ; it is required to.
Side 230 - To describe an isosceles triangle, having each of the angles at the base double of the third angle.
Side 157 - 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.
Side 33 - FC ; (ax. 1.) and FA, FB, 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 xxv - ... the squares of the diagonals, is equal to the sum of the squares of the bisected sides together with four times the square of the line joining those points of bisection.
Side 248 - If an equilateral triangle be inscribed in a circle, and the adjacent arcs cut off by two of its sides be bisected, the line joining the points of bisection shall be trisected by the sides.
Side 355 - The sine of an angle is equal to the sine of its supplement. The sine rule Consider fig.