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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Resultat 1-5 av 10
Side 40
... gnomon . Thus the paral- A ' lelogram HG , together with the com- ' plements AF , FC , is the gnomon , which is more briefly expressed by ' the letters AGK , or EHC , which are ' at the opposite angles of the paral- ' lelograms which ...
... gnomon . Thus the paral- A ' lelogram HG , together with the com- ' plements AF , FC , is the gnomon , which is more briefly expressed by ' the letters AGK , or EHC , which are ' at the opposite angles of the paral- ' lelograms which ...
Side 43
... gnomon CMG , together with LG , is equal to the rectangle AD , DB , together with the square of CD : but the gnomon CMG and LG makes up the whole figure CEFB , which is the square of CB : there- fore the rectangle AD , DB , together ...
... gnomon CMG , together with LG , is equal to the rectangle AD , DB , together with the square of CD : but the gnomon CMG and LG makes up the whole figure CEFB , which is the square of CB : there- fore the rectangle AD , DB , together ...
Side 44
... gnomon CMG : and DM is the rectangle contained by AD , DB , for DM is equal ( Cor . 4. 2. ) to DB : therefore the gnomon CMG is equal to the rectangle AD , DB , add to each of these LG , which is equal to the square of CB , therefore ...
... gnomon CMG : and DM is the rectangle contained by AD , DB , for DM is equal ( Cor . 4. 2. ) to DB : therefore the gnomon CMG is equal to the rectangle AD , DB , add to each of these LG , which is equal to the square of CB , therefore ...
Side 45
... gnomon AOH was demonstrated to be quadruple of AK : there- fore four times the rectangle AB , BC is equal to the gnomon AOH . To each of these add XH , which is equal ( Cor . 4. 2. ) to the square of AC : therefore four times the ...
... gnomon AOH was demonstrated to be quadruple of AK : there- fore four times the rectangle AB , BC is equal to the gnomon AOH . To each of these add XH , which is equal ( Cor . 4. 2. ) to the square of AC : therefore four times the ...
Side 146
... gnomon CHL : thereforce CE , or the parallelogram AD is greater , than the parallelogram AF . Next , let AK the base of AF be less than AC , and , the same construction being made , the parallelogram DH is equal to DG ( 36 . 1. ) , for ...
... gnomon CHL : thereforce CE , or the parallelogram AD is greater , than the parallelogram AF . Next , let AK the base of AF be less than AC , and , the same construction being made , the parallelogram DH is equal to DG ( 36 . 1. ) , for ...
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.