Euclid's Elements of plane geometry [book 1-6] explicitly enunciated, by J. Pryde. [With] Key1860 |
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Side 10
Euclides James Pryde. equal to it . Therefore , also , the whole triangle ABC shall coincide with the whole triangle DEF , and be equal to it ; and the remaining angles of the one shall coincide with the remaining angles of the other ...
Euclides James Pryde. equal to it . Therefore , also , the whole triangle ABC shall coincide with the whole triangle DEF , and be equal to it ; and the remaining angles of the one shall coincide with the remaining angles of the other ...
Side 11
... ABC a triangle having the angle ABC equal to the angle ACB , to prove that the side AB is also equal to the side AC . and join DC . but A ( Const . ) For , if AB be not equal to AC , one of them is greater than the other . Let AB be the ...
... ABC a triangle having the angle ABC equal to the angle ACB , to prove that the side AB is also equal to the side AC . and join DC . but A ( Const . ) For , if AB be not equal to AC , one of them is greater than the other . Let AB be the ...
Side 12
... triangle ACB ; produce AC , AD , to E , F ; then , because AC is equal to AD in the triangle ACD , the angles ECD ... ABC and DEF , two triangles having the two sides AB and AC , equal to the two sides DE and DF , each to each- namely , AB to ...
... triangle ACB ; produce AC , AD , to E , F ; then , because AC is equal to AD in the triangle ACD , the angles ECD ... ABC and DEF , two triangles having the two sides AB and AC , equal to the two sides DE and DF , each to each- namely , AB to ...
Side 13
... triangle ABC , and bisect ( I. 9 ) the angle ACB by the straight line CD . parts in the point D. AB is cut into two equal ( Dem . ) Because AC is equal to CB , and CD common to the two triangles ACD and BCD ; A C D the two sides AC and ...
... triangle ABC , and bisect ( I. 9 ) the angle ACB by the straight line CD . parts in the point D. AB is cut into two equal ( Dem . ) Because AC is equal to CB , and CD common to the two triangles ACD and BCD ; A C D the two sides AC and ...
Side 16
... ABC and ABD are likewise together equal to two right angles ; therefore the angles CBA and ABE are equal to the ... triangle be produced , the exterior angle is greater than either of the interior opposite angles . Given a triangle ABC ...
... ABC and ABD are likewise together equal to two right angles ; therefore the angles CBA and ABE are equal to the ... triangle be produced , the exterior angle is greater than either of the interior opposite angles . Given a triangle ABC ...
Andre utgaver - Vis alle
Euclid's Elements of plane geometry [book 1-6] explicitly enunciated, by J ... Euclides Uten tilgangsbegrensning - 1860 |
Euclid's Elements of Plane Geometry [Book 1-6] Explicitly Enunciated, by J ... Euclides,James Pryde Ingen forhåndsvisning tilgjengelig - 2023 |
Euclid's Elements of Plane Geometry [book 1-6] Explicitly Enunciated, by J ... Euclides,James Pryde Ingen forhåndsvisning tilgjengelig - 2018 |
Vanlige uttrykk og setninger
ABCD adjacent angles angle ABC angle ACB angle BAC apothem base BC BC is equal bisected centre Chambers's chord circle ABC circumference Const cosec cosine described diameter divided double draw equal angles equal to twice equiangular equilateral equilateral polygon equimultiples exterior angle fore given line given point given straight line gnomon greater hence hypotenuse inscribed isosceles triangle less line drawn multiple number of sides opposite angle parallel parallelogram perimeter perpendicular polygon produced proportional PROPOSITION prove radius ratio rectangle contained rectilineal figure regular polygon remaining angle right angles right-angled triangle segment semiperimeter shewn similar sine square on AC straight line AC tangent THEOREM third touches the circle triangle ABC triangle DEF twice the rectangle vertical angle wherefore
Populære avsnitt
Side 23 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz. either the sides adjacent to the equal...
Side 52 - 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, together with the square of half the line bisected, is equal to the square of the straight line which is made up of the half and the part produced.
Side 51 - 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.
Side 53 - 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 in the point C ; the squares of AB, BC are equal to twice the rectangle AB, BC...
Side 3 - When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a right angle ; and the straight line which stands on the other is called a perpendicular to it.
Side 29 - 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.
Side 117 - And the same thing is to be understood when it is more briefly expressed by saying, a has to d the ratio compounded of the ratios of e to f, g to h, and k to l. In like manner, the same things being supposed, if m has to n the same ratio which a has to d ', then, for shortness...
Side 13 - Upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity.
Side 159 - From the point A draw a straight line AC, making any angle with AB ; and in AC take any point D, and take AC the same multiple of AD, that AB is of the part which is to be cut off from it : join BC, and draw DE parallel to it : then AE is the part required to be cut off. Because ED is parallel to one of the sides of the triangle ABC, viz. to BC ; as CD is to DA, so is (2.
Side 60 - CB, BA, by twice the rectangle CB, BD. Secondly, Let AD fall without the triangle ABC. Then, because the angle at D is a right angle, the angle ACB is greater than a right angle ; (i.