The geometry, by T. S. Davies. Conic sections, by Stephen FenwickJ. Weale, 1853 |
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Αποτελέσματα 1 - 5 από τα 47.
Σελίδα 14
... reason , the exterior angle CEB of the triangle ABE is greater than BAC ; And it has been demonstrated that the angle BDC is greater than the angle CEB ; much more then is the angle BDC greater than the angle BAC . Therefore , if from ...
... reason , the exterior angle CEB of the triangle ABE is greater than BAC ; And it has been demonstrated that the angle BDC is greater than the angle CEB ; much more then is the angle BDC greater than the angle BAC . Therefore , if from ...
Σελίδα 22
... reason EF is equal to BC ; wherefore AD is equal ( 1. Ax . ) to EF ; and DE is common ; therefore the whole , or the re- mainder AE , is equal ( 2 or 3 Ax . ) to the whole , or the remainder DF : AB also is equal to DC ; and the two EA ...
... reason EF is equal to BC ; wherefore AD is equal ( 1. Ax . ) to EF ; and DE is common ; therefore the whole , or the re- mainder AE , is equal ( 2 or 3 Ax . ) to the whole , or the remainder DF : AB also is equal to DC ; and the two EA ...
Σελίδα 23
... reason , the parallelogram EFGH is equal to the same EBCH : Therefore also the parallelogram ABCD is equal ( 1 Ax . ) to EFGH . Wherefore parallelograms , etc. Q. E. D. PROPOSITION XXXVII . THEOR . Triangles upon the same base , and ...
... reason , the parallelogram EFGH is equal to the same EBCH : Therefore also the parallelogram ABCD is equal ( 1 Ax . ) to EFGH . Wherefore parallelograms , etc. Q. E. D. PROPOSITION XXXVII . THEOR . Triangles upon the same base , and ...
Σελίδα 26
... reason , the triangle KGC is equal to the triangle KFC : B K Then , because the triangle AEK is equal to the triangle AHK , and the triangle KGC to KFC ; the triangle AEK , together with the triangle KGC , is equal ( 2 Ax . ) to the ...
... reason , the triangle KGC is equal to the triangle KFC : B K Then , because the triangle AEK is equal to the triangle AHK , and the triangle KGC to KFC ; the triangle AEK , together with the triangle KGC , is equal ( 2 Ax . ) to the ...
Σελίδα 28
... to two right angles ; therefore CA is in the same straight line ( 14. 1. ) with AG ; For the same reason , AB and AH are in the same straight line ; E L And because the angle DBC is equal ( 11 Ax 28 EUCLID'S ELEMENTS .
... to two right angles ; therefore CA is in the same straight line ( 14. 1. ) with AG ; For the same reason , AB and AH are in the same straight line ; E L And because the angle DBC is equal ( 11 Ax 28 EUCLID'S ELEMENTS .
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Συχνά εμφανιζόμενοι όροι και φράσεις
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
Δημοφιλή αποσπάσματα
Σελίδα 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.
Σελίδα 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.
Σελίδα 4 - AB; but things which are equal to the same are equal to one another...
Σελίδα 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.
Σελίδα 8 - If two triangles have two sides of the one equal to two sides of the...
Σελίδα 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...
Σελίδα 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.
Σελίδα 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.
Σελίδα 4 - Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another.
Σελίδα 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.