The geometry, by T. S. Davies. Conic sections, by Stephen FenwickJ. Weale, 1853 |
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Página 4
... proved that CA is equal to AB ; therefore CA , CB are each of them equal to AB : But things which are equal to the same are equal to one another ( 1 Axiom ) ; therefore CA is equal to CB ; wherefore CA , AB , BC are equal to one another ...
... proved that CA is equal to AB ; therefore CA , CB are each of them equal to AB : But things which are equal to the same are equal to one another ( 1 Axiom ) ; therefore CA is equal to CB ; wherefore CA , AB , BC are equal to one another ...
Página 6
... proved to be equal to GB ; therefore the two sides BF , FC are equal to the two CG , GB , each to each ; and the angle BFC is equal to the angle CGB ; wherefore the triangles BFC , CGB are equal ( 4. 6 EUCLID'S ELEMENTS .
... proved to be equal to GB ; therefore the two sides BF , FC are equal to the two CG , GB , each to each ; and the angle BFC is equal to the angle CGB ; wherefore the triangles BFC , CGB are equal ( 4. 6 EUCLID'S ELEMENTS .
Página 7
... proved that the angle FBC is equal to the angle GCB , which are the angles upon the other side of the base . Therefore the angles at the base , etc. Q. E. D. COROLLARY . Hence every equilateral triangle is also equiangular . PROPOSITION ...
... proved that the angle FBC is equal to the angle GCB , which are the angles upon the other side of the base . Therefore the angles at the base , etc. Q. E. D. COROLLARY . Hence every equilateral triangle is also equiangular . PROPOSITION ...
Página 8
... proved to be greater than the same BCD ; which is impossible . The case in which the vertex of one triangle is upon a side of the other , needs no demon- stration . A B Therefore , upon the same base , and on the same side of it , there ...
... proved to be greater than the same BCD ; which is impossible . The case in which the vertex of one triangle is upon a side of the other , needs no demon- stration . A B Therefore , upon the same base , and on the same side of it , there ...
Página 30
... Prove the second corollary of i . 32 independently of the first , and from it deduce the first . Also , ( a ) Show that each angle of an equilateral triangle is two - thirds of a right angle ; ( b ) That the two acute angles of a right ...
... Prove the second corollary of i . 32 independently of the first , and from it deduce the first . Also , ( a ) Show that each angle of an equilateral triangle is two - thirds of a right angle ; ( b ) That the two acute angles of a right ...
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Términos y frases comunes
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
Pasajes populares
Página 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.
Página 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.
Página 4 - AB; but things which are equal to the same are equal to one another...
Página 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.
Página 8 - If two triangles have two sides of the one equal to two sides of the...
Página 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...
Página 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.
Página 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.
Página 4 - Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another.
Página 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.