Elements of Plane Geometry According to EuclidW. and R. Chambers, 1837 - 240 páginas |
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Página 6
... reason- ing from the premises to the conclusion . 10. An indirect demonstration establishes a proposition , by proving that any hypothesis contrary to it , is contradic- tory or absurd ; and it is , therefore , sometimes called a ...
... reason- ing from the premises to the conclusion . 10. An indirect demonstration establishes a proposition , by proving that any hypothesis contrary to it , is contradic- tory or absurd ; and it is , therefore , sometimes called a ...
Página 20
... 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 ; still greater , then , is the angle BDC than the angle BAC . PROPOSITION XXII ...
... 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 ; still greater , then , is the angle BDC than the angle BAC . PROPOSITION XXII ...
Página 31
... reason EF is equal to BC ; where- fore AD is equal to EF ( Ax . 1 ) ; and DE is common ; therefore ( Ax . 2 or 3 ) the whole or the remainder AE is equal to the whole or the remainder DF ; AB also is equal to DC ; and the two EA , AB ...
... reason EF is equal to BC ; where- fore AD is equal to EF ( Ax . 1 ) ; and DE is common ; therefore ( Ax . 2 or 3 ) the whole or the remainder AE is equal to the whole or the remainder DF ; AB also is equal to DC ; and the two EA , AB ...
Página 32
... reason , the parallelogram EFGH is equal to the same EBCH ; therefore also the parallelogram ABCD is equal to EFGH.- PROPOSITION XXXVII . THEOREM . Triangles upon the same base , and between the same pa- rallels , are equal to one ...
... reason , the parallelogram EFGH is equal to the same EBCH ; therefore also the parallelogram ABCD is equal to EFGH.- PROPOSITION XXXVII . THEOREM . Triangles upon the same base , and between the same pa- rallels , are equal to one ...
Página 37
... reason , the triangle KGC is equal to the triangle KFC . Then , because the triangle AĒK is equal to the triangle AHK , and the triangle KGC to KFC ; the triangle AEK , together with the triangle KGC , is equal to the triangle AHK ...
... reason , the triangle KGC is equal to the triangle KFC . Then , because the triangle AĒK is equal to the triangle AHK , and the triangle KGC to KFC ; the triangle AEK , together with the triangle KGC , is equal to the triangle AHK ...
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Elements of Plane Geometry According to Euclid Robert Simson,Formerly Chairman Department of Immunology John Playfair,John Playfair Sin vista previa disponible - 2016 |
Términos y frases comunes
ABCD AC is equal angle ABC angle ACB angle BAC angle BCD angle EDF apothem base BC bisected centre chord circle ABC circumference described diameter double draw equal angles equal to AC equiangular equilateral polygon equimultiples exterior angle fore geometry given circle given line given point given rectilineal given straight line gnomon greater hypotenuse inscribed interminate less Let ABC magnitudes multiple opposite angle parallel parallelogram perimeter perpendicular polygon porism produced proportional PROPOSITION radius rectangle AB BC rectangle contained rectilineal figure regular polygon remaining angle right angles right-angled triangle Schol segment semicircle semiperimeter similar sine square of AC tangent THEOREM touches the circle triangle ABC triangle DEF twice the rectangle vulgar fraction wherefore
Pasajes populares
Página 1 - Things which are equal to the same thing are equal to one another. 2. If equals be added to equals the wholes are equal. 3. If equals be taken from equals the remainders are equal. 4. If equals be added to unequals the wholes are unequal. 5. If equals be taken from unequals the remainders are unequal. 6. Things which are double of the same thing are equal to one another.
Página 73 - The angle in a semicircle is a right angle; the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle.
Página 9 - To bisect a given finite straight line, that is, to divide it into two equal parts. Let AB be the given straight line : it is required to divide it intotwo equal parts.
Página 4 - If two triangles have two sides of the one equal to two sides of the...
Página 139 - 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 BC, CG, be placed in a straight line ; therefore DC and CE are also in a straight line (2.
Página 23 - Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Página 129 - 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.
Página 80 - A circle is said to be described about a rectilineal figure, when the circumference of the circle passes through all the angular points of the figure about which it is described. 7. A straight line is said to be placed in a circle, when the extremities of it are in the circumference of the circle.
Página 27 - Parallelograms upon equal bases, and between the same parallels, are equal to one another.
Página 44 - If a straight line be divided into any two parts, the squares of the whole line and of one of the parts are equal to twice the rectangle contained by the whole and that part, together with the square of the other part. Let the straight line AB be divided into any two parts at the point C : the squares of AB, BC shall be equal to twice the rectangle AB, BC, together with the square of AC.