Elements of Geometry, Containing the First Six Books of EuclidBaldwin, Cradock, and Joy, 1826 - 180 sider |
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Side 5
... consequently , the triangle ABC is equilateral , and it is described upon the given finite right line AB . Q. E. F. PROPOSITION II . PROBLEM . From a given point to draw a right line equal to a given right line . F Let A be the given ...
... consequently , the triangle ABC is equilateral , and it is described upon the given finite right line AB . Q. E. F. PROPOSITION II . PROBLEM . From a given point to draw a right line equal to a given right line . F Let A be the given ...
Side 13
... consequently the finite right line AB is bisected in the point D. Q. E. F. BD . Deduction . From the vertex of a given scalene triangle , to draw , to the base , a straight line which shall exceed the less of the two sides , as much as ...
... consequently the finite right line AB is bisected in the point D. Q. E. F. BD . Deduction . From the vertex of a given scalene triangle , to draw , to the base , a straight line which shall exceed the less of the two sides , as much as ...
Side 41
... consequently ABDE is a rectangle . But it was shown to be equilateral , there- fore it is a square , and has been described on the given right line AB . Q. E. F. COROLLARY . Hence every parallelogram having one right angle is a ...
... consequently ABDE is a rectangle . But it was shown to be equilateral , there- fore it is a square , and has been described on the given right line AB . Q. E. F. COROLLARY . Hence every parallelogram having one right angle is a ...
Side 75
... consequently AB is double of For the same reason CD is also double of co , and AB is equal to CD ; therefore also AF is equal to CG , and because AE is equal to EC , the square of AE will be equal to the square of but the EC , squares ...
... consequently AB is double of For the same reason CD is also double of co , and AB is equal to CD ; therefore also AF is equal to CG , and because AE is equal to EC , the square of AE will be equal to the square of but the EC , squares ...
Side 79
... consequently the angle FGC is greater than the angle FCG . But the D greater side subtends the greater angle.b B GE Wherefore FC is greater than FG . But FC is equal to FB : wherefore FB is greater than FG , the less than the greater ...
... consequently the angle FGC is greater than the angle FCG . But the D greater side subtends the greater angle.b B GE Wherefore FC is greater than FG . But FC is equal to FB : wherefore FB is greater than FG , the less than the greater ...
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Elements of Geometry: Containing the First Six Books of Euclid: With a ... John Playfair Uten tilgangsbegrensning - 1855 |
Vanlige uttrykk og setninger
ABC is equal adjacent angles Algebra angle ABC angle ACB angle BAC angles equal base BC bisected centre circle ABC circum circumference BC diameter double draw equal angles equal circles equal right lines equal to F equi equiangular equimultiples Euclid EUCLID'S ELEMENTS exceed exterior angle fore four magnitudes fourth Geometry given circle given point given right line gnomon greater ratio hence inscribed join less Let ABC multiple parallel parallelogram perpendicular polygon proportional Q. E. D. Deduction Q. E. D. PROPOSITION rectangle contained remaining angle right angles right line AB right line AC sector HEF segment side BC similar and similarly square of AC subtending THEOREM tiple touches the circle triangle ABC triangle DEF whence whole
Populære avsnitt
Side xxvi - A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference. XVIII. A semicircle is the figure contained by a diameter and the part of the circumference cut off by the diameter. XIX. "A segment of a circle is the figure contained by a straight line, and the circumference it cuts off.
Side 74 - The straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls without the circle...
Side 33 - The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another.
Side 148 - 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.
Side 27 - And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, therefore the whole angle ABD is equal to the whole angle ACD • (ax.
Side 8 - 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.
Side 73 - DH; (I. def. 15.) therefore DH is greater than DG, the less than the greater, which is impossible : therefore no straight line can be drawn from the point A, between AE and the circumference, which does not cut the circle : or, which amounts to the same thing, however great an acute angle a straight line makes with the diameter at the point A, or however small an angle it makes with AE, the circumference must pass between that straight line and the perpendicular AE.
Side 99 - To describe a square about a given circle. Let ABCD be the given circle ; it is required to describe a square about it. . Draw two diameters AC, BD of the circle ABCD, at right angles to one another, and through the points A, B, • 17.3. C, D, draw...
Side 7 - ... equal to them, of the other. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB equal to DE, and AC to DF ; but the base CB greater than the base EF ; the angle BAC is likewise greater than the angle EDF.
Side 88 - From a given circle to cut off a segment, which shall contain an angle equal to a given rectilineal angle.