Euclid's Elements: Or, Second Lessons in Geometry,in the Order of Simson's and Playfair's Editions ...Collins, Brother & Company, 1846 - 138 σελίδες |
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Σελίδα 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 ...
Σελίδα 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 ...
Σελίδα 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 ...
Σελίδα 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 ...
Σελίδα 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 ...
Άλλες εκδόσεις - Προβολή όλων
Euclid's Elements, Or Second Lessons in Geometry, in the Order of Simson's ... D. M'Curdy Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2017 |
Euclid's Elements, Or Second Lessons in Geometry, in the Order of Simson's ... D. M'Curdy Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2017 |
Συχνά εμφανιζόμενοι όροι και φράσεις
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
Δημοφιλή αποσπάσματα
Σελίδα 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.
Σελίδα 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.
Σελίδα 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.
Σελίδα 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.
Σελίδα 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...
Σελίδα 26 - Triangles upon equal bases, and between the same parallels, are equal to one another.
Σελίδα 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.
Σελίδα 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.
Σελίδα 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.
Σελίδα 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.