The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago Vitiated These Books are Corrected, and Some of Euclid's Demonstrations are Restored. Also, the Book of Euclid's Data, in Like Manner Corrected. viz. the first six books, together with the eleventh and twelfthDesilver, Thomas & Company, 1838 - 416 sider |
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Side 3
... demonstrated , and not assumed ; and therefore , though this were a true Proposition , it ought to have been demonstrated . But , in- deed , this Proposition , which makes the 10th Definition of the 11th Book , is not true universally ...
... demonstrated , and not assumed ; and therefore , though this were a true Proposition , it ought to have been demonstrated . But , in- deed , this Proposition , which makes the 10th Definition of the 11th Book , is not true universally ...
Side 12
... demonstrated . PROP . V. THEOR . 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 . Let ABC be an isosceles ...
... demonstrated . PROP . V. THEOR . 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 . Let ABC be an isosceles ...
Side 13
... demonstrated , that the whole angle ABG is equal to the whole ACF , the parts of which , the angles CBG , BCF are also equal ; the remaining angle ABC is therefore equal to the remaining angle ACB , which are the angles at the base of ...
... demonstrated , that the whole angle ABG is equal to the whole ACF , the parts of which , the angles CBG , BCF are also equal ; the remaining angle ABC is therefore equal to the remaining angle ACB , which are the angles at the base of ...
Side 14
... demonstrated to be greater than it ; which is impossible . A C C D D E F B But if one of the vertices , as D , be within the other triangle ACB ; produce AC , AD to E , F ; there- fore , because AC is equal to AD in the triangle ACD ...
... demonstrated to be greater than it ; which is impossible . A C C D D E F B But if one of the vertices , as D , be within the other triangle ACB ; produce AC , AD to E , F ; there- fore , because AC is equal to AD in the triangle ACD ...
Side 16
... demonstrated , that two straight lines cannot have a common segment . E If it be possible , let the two straight lines ABC , ABD have the seg- ment AB common to both of them . From the point B draw BE at right angles to AB ; and because ...
... demonstrated , that two straight lines cannot have a common segment . E If it be possible , let the two straight lines ABC , ABD have the seg- ment AB common to both of them . From the point B draw BE at right angles to AB ; and because ...
Vanlige uttrykk og setninger
altitude angle ABC angle BAC base BC BC is equal BC is given bisected centre circle ABCD circumference Co-S cone cylinder demonstrated described diameter draw drawn equal angles equiangular equimultiples Euclid ex æquali excess fore given angle given in magnitude given in position given in species given magnitude given ratio given straight line gles gnomon join less Let ABC multiple parallel parallelogram parallelogram AC perpendicular point F polygon prism proportionals proposition pyramid Q. E. D. PROP radius ratio of AE rectangle CB rectangle contained rectilineal figure remaining angle right angles segment sides BA similar sine solid angle solid parallelopiped square of AC square of BC straight line AB straight line BC tangent THEOR third triangle ABC triplicate ratio vertex wherefore
Populære avsnitt
Side 45 - Ir a straight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the square of the other part, is equal to the square of the straight line which is made up of the whole and that part.
Side 41 - If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part.
Side 54 - Ir any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle. Let ABC be a circle, and A, B any two points in the circumference ; the straight line drawn from A to B shall fall within the circle.
Side 18 - ABD, the less to the greater, which is impossible ; therefore BE is not in the same straight line with BC.
Side 10 - From a given point to draw a straight line equal to a given straight line. Let A be the given point, and BC the given straight line: it is required to draw from the point A a straight line equal to BC.
Side 8 - Let it be granted that a straight line may be drawn from any one point to any other point.
Side 256 - Again ; the mathematical postulate, that " things which are equal to the same are equal to one another," is similar to the form of the syllogism in logic, which unites things agreeing in the middle term.
Side 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.
Side 23 - At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle. Let AB be the given straight line, and A...
Side 20 - ANY two angles of a triangle are together less than two right angles.