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 ... |
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Page 5
... BC two straight lines equal to each other , viz . one on each side of AD . Make DE = DF and join AE . Then AE = AF ... parallel to BD , and .. perpendicular to EF . The triangles ACE , DFC , having each Sect . 1.1 ] 5 GEOMETRICAL PROBLEMS .
... BC two straight lines equal to each other , viz . one on each side of AD . Make DE = DF and join AE . Then AE = AF ... parallel to BD , and .. perpendicular to EF . The triangles ACE , DFC , having each Sect . 1.1 ] 5 GEOMETRICAL PROBLEMS .
Page 7
... BC in the given ratio . Join AC , and from P draw PDE parallel to AC . PDE is the line required . P A D 10- P For since DE is parallel to AC , ( Eucl . vi . 2. ) DB : BE :: AB : BC , i . e . in the given ratio . ( 12. ) If from a given ...
... BC in the given ratio . Join AC , and from P draw PDE parallel to AC . PDE is the line required . P A D 10- P For since DE is parallel to AC , ( Eucl . vi . 2. ) DB : BE :: AB : BC , i . e . in the given ratio . ( 12. ) If from a given ...
Page 11
... parallel to BC ; and therefore parallel to each other ; and draw DK parallel to AB . Then because GD is parallel to HE one of the sides of the triangle AHE , AG : GH :: AD : DE ; hence AG = GH . For the same reason DL = LM . But DM being ...
... parallel to BC ; and therefore parallel to each other ; and draw DK parallel to AB . Then because GD is parallel to HE one of the sides of the triangle AHE , AG : GH :: AD : DE ; hence AG = GH . For the same reason DL = LM . But DM being ...
Page 12
... BC is parallel to FD , the angle BCG is equal to GDF and the vertically opposite angles at G are equal ; therefore the triangles DGF , BGC are similar , and BC : BG :: FD : FG But FE being parallel to BC , ( Eucl . vi . 2. ) AB : BC ...
... BC is parallel to FD , the angle BCG is equal to GDF and the vertically opposite angles at G are equal ; therefore the triangles DGF , BGC are similar , and BC : BG :: FD : FG But FE being parallel to BC , ( Eucl . vi . 2. ) AB : BC ...
Page 16
... line , may be equal to a given line . Let AB , BC be the lines given in position , and A the D C H K B I E given point . Draw AD perpendicular to AB , and meet- ing BC in D ; draw DE parallel to AB and equal to the given line . And draw EF ...
... line , may be equal to a given line . Let AB , BC be the lines given in position , and A the D C H K B I E given point . Draw AD perpendicular to AB , and meet- ing BC in D ; draw DE parallel to AB and equal to the given line . And draw EF ...
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Common terms and phrases
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
Popular passages
Page 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.
Page 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.
Page 230 - To describe an isosceles triangle, having each of the angles at the base double of the third angle.
Page 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.
Page 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...
Page 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.
Page 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.
Page 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.
Page 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.
Page 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.