The geometry, by T. S. Davies. Conic sections, by Stephen FenwickJ. Weale, 1853 - Mathematics |
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Page 6
... common to the two triangles AFC , AGB ; therefore the base FC is equal ( 4. 1. ) to the base GB , and the triangle AFC to the triangle AGB ; and the remaining angles of the one are equal ( 4. 1. ) to the remaining angles of the other ...
... common to the two triangles AFC , AGB ; therefore the base FC is equal ( 4. 1. ) to the base GB , and the triangle AFC to the triangle AGB ; and the remaining angles of the one are equal ( 4. 1. ) to the remaining angles of the other ...
Page 7
... common to both , the two sides , DB , BC are equal to the two AC , CB , each to each ; and the angle DBC is equal to the angle ( Hyp . ) ACB ; therefore the base DC is equal to the base AB , and the triangle DBC is equal to the triangle ...
... common to both , the two sides , DB , BC are equal to the two AC , CB , each to each ; and the angle DBC is equal to the angle ( Hyp . ) ACB ; therefore the base DC is equal to the base AB , and the triangle DBC is equal to the triangle ...
Page 9
... common to the two triangles DAF , EAF ; the two sides DA , AF , are equal to the two sides EA , AF , each to each ; and the base DF is equal to the base EF ( 24 Def . ) ; therefore the angle DAF is equal ( 8. 1. ) to the angle EAF ...
... common to the two triangles DAF , EAF ; the two sides DA , AF , are equal to the two sides EA , AF , each to each ; and the base DF is equal to the base EF ( 24 Def . ) ; therefore the angle DAF is equal ( 8. 1. ) to the angle EAF ...
Page 10
... common segment . E If it be possible , let the two straight lines ABC , ABD have the segment AB common to both of them . From the point B draw BE at right angles to AB ; and because ABC is a straight line , the angle CBE is equal ( 10 ...
... common segment . E If it be possible , let the two straight lines ABC , ABD have the segment AB common to both of them . From the point B draw BE at right angles to AB ; and because ABC is a straight line , the angle CBE is equal ( 10 ...
Page 11
... common angle ABC , the remaining angle ABE is equal ( 3 Ax . ) to the remaining angle ABD , the less to the greater , which is impossible ; therefore BE is not in the same straight line with BC . And , in like manner , it may he ...
... common angle ABC , the remaining angle ABE is equal ( 3 Ax . ) to the remaining angle ABD , the less to the greater , which is impossible ; therefore BE is not in the same straight line with BC . And , in like manner , it may he ...
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Common terms and phrases
ABCD axis base bisected called centre circle circumference coincide common cone construction contained coordinate curve described diameter difference dihedral angles direction distance divided double draw drawn edges ellipse equal equal angles equimultiples extremities faces figure formed four fourth given line given point greater hence horizontal inclination intersection join less likewise magnitudes manner meet method multiple opposite parallel parallelogram pass perpendicular perspective picture plane MN plane of projection position preceding prisms problem produced projector Prop proportional PROPOSITION proved ratio reason rectangle remaining respectively right angles SCHOLIUM segment shown sides similar sphere square straight line surface taken tangent THEOR third touch trace transverse triangle triangle ABC trihedral vertex vertical Whence Wherefore whole
Popular passages
Page 19 - That, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.
Page 35 - If a straight line be divided into two equal parts, and also into two unequal parts ; the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line.
Page 4 - AB; but things which are equal to the same are equal to one another...
Page 128 - EQUIANGULAR parallelograms have to one another the ratio which is compounded of the ratios of their sides.* Let AC, CF be equiangular parallelograms, having the angle BCD equal to the angle ECG : the ratio of the parallelogram AC to the parallelogram CF, is the same with the ratio which is compounded of the ratios of their sides. Let BG, CG, be placed in a straight line ; therefore DC and CE are also in a straight line (14.
Page 8 - If two triangles have two sides of the one equal to two sides of the...
Page 36 - If a straight line be bisected and produced to any point, the rectangle contained by the whole line thus produced and the part of it produced...
Page 21 - BCD, and the other angles to the other angles, (4. 1.) each to each, to which the equal sides are opposite : therefore the angle ACB is equal to the angle CBD ; and because the straight line BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD equal to one another, AC is parallel (27. 1 .) to DB ; and it was shown to be equal to it. Therefore straight lines, &c.
Page 65 - If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles which this line makes with the line touching the circle shall be equal to the angles which are in the alternate segments of the circle.
Page 4 - Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another.
Page 116 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.