Euclid's Elements: Or, Second Lessons in Geometry,in the Order of Simson's and Playfair's Editions ...Collins, Brother & Company, 1846 - 138 páginas |
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Página 8
... half the compass of the angular point . Cor . When the declination of two straight lines is equal to half the compass of the point in which they meet , they form no angle , but are in one straight line . 12. A plane figure is any form ...
... half the compass of the angular point . Cor . When the declination of two straight lines is equal to half the compass of the point in which they meet , they form no angle , but are in one straight line . 12. A plane figure is any form ...
Página 25
... half of the parallelogram , which is bisected by BC . Wherefore , the opposite sides and angles , & c . , Recite ( a ) , p . 9 , 10 ; ( d ) , p . 27 ; ( b ) , def . above ; ( e ) , p . 32 ; Q. E. D. ( c ) , def . 16 ; ( f ) , p . 26 ...
... half of the parallelogram , which is bisected by BC . Wherefore , the opposite sides and angles , & c . , Recite ( a ) , p . 9 , 10 ; ( d ) , p . 27 ; ( b ) , def . above ; ( e ) , p . 32 ; Q. E. D. ( c ) , def . 16 ; ( f ) , p . 26 ...
Página 26
... half the parallelogram DBCF , because the diameter DC bi- sects it ; and the halves of equals are equal ( d ) ... half the parallel- ogram DEFH ; and the triangle ABC is half the parallelogram GBCA ( c ) : because they are bisected by the ...
... half the parallelogram DBCF , because the diameter DC bi- sects it ; and the halves of equals are equal ( d ) ... half the parallel- ogram DEFH ; and the triangle ABC is half the parallelogram GBCA ( c ) : because they are bisected by the ...
Página 29
... half the base of the given triangle C , make a parallelogram GBEF , equal to C ( a ) ; and let its angle GBE equal the given angle D ( b ) ; let AB , the given straight line , be the produced part of EB ; through B produce GB to M ...
... half the base of the given triangle C , make a parallelogram GBEF , equal to C ( a ) ; and let its angle GBE equal the given angle D ( b ) ; let AB , the given straight line , be the produced part of EB ; through B produce GB to M ...
Página 30
... half : C E H Also , the triangle ABD is upon the side BA , and between the same parallels as the parallelogram BL , of which it is equal to the half ( d ) . But the triangles BCF , ABD are equal ; having two sides BF , BC , in the one ...
... half : C E H Also , the triangle ABD is upon the side BA , and between the same parallels as the parallelogram BL , of which it is equal to the half ( d ) . But the triangles BCF , ABD are equal ; having two sides BF , BC , in the one ...
Otras ediciones - Ver todas
Euclid's Elements, Or Second Lessons in Geometry, in the Order of Simson's ... D. M'Curdy Sin vista previa disponible - 2017 |
Euclid's Elements, Or Second Lessons in Geometry, in the Order of Simson's ... D. M'Curdy Sin vista previa disponible - 2017 |
Términos y frases comunes
ABCD alternate angles angle ACD angles ABC angles equal antecedents Argument base BC bisected centre Chart chord circle ABC circumference Constr Denison Olmsted diameter draw drawn equal angles equal arcs equal radii equal sides equals the squares equi equiangular equilateral equilateral polygon equimultiples exterior angle fore Geometry given circle given rectilineal given straight line gles gnomon greater half inscribed isosceles isosceles triangle join less meet multiple opposite angles parallelogram parallelopipeds pentagon perimeter perpendicular plane polygon produced propositions Q. E. D. Recite radius ratio rectangle rectilineal figure School segment semicircle similar similar triangles sine square of AC tangent touches the circle triangle ABC unequal Wherefore
Pasajes populares
Página 90 - If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, the triangles shall be equiangular, and shall have those angles equal which are opposite to the homologous sides.
Página 117 - In the same way it may be proved that a : b : : sin. A : sin. B, and these two proportions may be written a : 6 : c : : sin. A : sin. B : sin. C. THEOREM III. t8. In any plane triangle, the sum of any two sides is to their difference as the tangent of half the sum of the opposite angles is to the tangent of half their difference. By Theorem II. we have a : b : : sin. A : sin. B.
Página 92 - IN a right-angled triangle, if a perpendicular be drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle, and to one another.
Página 79 - THEOREM. lf the first has to the second the same ratio which the third has to the fourth, but the third to the fourth, a greater ratio than the fifth has to the sixth ; the first shall also have to the second a greater ratio than the fifth, has to the sixth.
Página 87 - 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...
Página 26 - Triangles upon equal bases, and between the same parallels, are equal to one another.
Página 94 - Equal parallelograms which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional ; and parallelograms that have one angle of the one equal to one angle of the other, and their sides about the equal angles reciprocally proportional, are equal to one another.
Página 12 - THE angles at the base of an isosceles triangle are equal to one another : and, if the equal sides be produced, the angles upon the other side of the base shall be equal.
Página 133 - If a straight line stand at right angles to each of two straight lines at the point of their intersection, it shall also be at right angles to the plane which passes through them, that is, to the plane in which they are.
Página 13 - AB be the greater, and from it cut (3. 1.) off DB equal to AC the less, and join DC ; therefore, because A in the triangles DBC, ACB, DB is equal to AC, and BC 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 ACB; therefore the base DC is equal to the base AB, and the triangle DBC is< equal to the triangle (4. 1.) ACB, the less to 'the greater; which is absurd.