The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh and Twelfth ... Also the Book of Euclid's Data, in Like Manner CorrectedWingrave and Collingwood, 1816 - 528 sider |
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Resultater 1-5 af 14
Side 43
... Gnomon . Thus ' the parallelogram HG , ' together with the com- ' plements AF , FC , is the gnomon , which is more ' briefly expressed by the " letters AGK , or EHC , " H F K В G ' which are at the opposite angles of the parallelograms ...
... Gnomon . Thus ' the parallelogram HG , ' together with the com- ' plements AF , FC , is the gnomon , which is more ' briefly expressed by the " letters AGK , or EHC , " H F K В G ' which are at the opposite angles of the parallelograms ...
Side 47
... gnomon CMG ; therefore the gnomon CMG is equal to the rectangle AD , DB : To each of these add LG , which is equal to the square of CD ; therefore the gnomon CMG , together with LG , is equal to the rectangle AD , DB , together with the ...
... gnomon CMG ; therefore the gnomon CMG is equal to the rectangle AD , DB : To each of these add LG , which is equal to the square of CD ; therefore the gnomon CMG , together with LG , is equal to the rectangle AD , DB , together with the ...
Side 48
... gnomon CMG : And AM L H M E G Cor . 4. 2. is the rectangle contained by AD , DB , for DM is equale to DB : Therefore the gnomon CMG is equal to the rect- angle AD , DB : Add to each of these LG , which is equal to the square of CB ...
... gnomon CMG : And AM L H M E G Cor . 4. 2. is the rectangle contained by AD , DB , for DM is equale to DB : Therefore the gnomon CMG is equal to the rect- angle AD , DB : Add to each of these LG , which is equal to the square of CB ...
Side 49
... gnomon AKF , together with the square CK ; therefore the gnomon AKF , toge- ther with the square CK , is double of AK : But twice the rectangle AB , BC is double of AK , for BK is equal to BC : Therefore the gno- mon AKF , together with ...
... gnomon AKF , together with the square CK ; therefore the gnomon AKF , toge- ther with the square CK , is double of AK : But twice the rectangle AB , BC is double of AK , for BK is equal to BC : Therefore the gno- mon AKF , together with ...
Side 50
... gnomon AOH was demonstrated to be quadruple of AK ; therefore four times the rectangle AB , BC , is equal to the gnomon AOH . To Cor . 4. 2. each of these add XH , which is equal to the square of AC : Therefore four times the rectangle ...
... gnomon AOH was demonstrated to be quadruple of AK ; therefore four times the rectangle AB , BC , is equal to the gnomon AOH . To Cor . 4. 2. each of these add XH , which is equal to the square of AC : Therefore four times the rectangle ...
Almindelige termer og sætninger
ABC is given ABCD AC is equal altitude angle ABC angle BAC base BC bisected Book XI centre circle ABC circumference common logarithm cone cylinder demonstrated described diameter draw drawn equal angles equiangular equimultiples Euclid excess fore given angle given in magnitude given in position given in species given magnitude given point given ratio given straight line gnomon greater join less Let ABC logarithm multiple opposite parallel parallelogram AC perpendicular point F polygon prism proportionals proposition 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 parallelopipeds square of BC straight line AB straight line BC tangent THEOR third triangle ABC vertex wherefore
Populære passager
Side 41 - If a straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts.
Side 180 - Wherefore, in equal circles &c. QED PROPOSITION B. THEOREM If the vertical angle of a triangle be bisected by a straight line which likewise cuts the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square on the straight line which bisects the angle.
Side 166 - 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 •f the ratios of their sides. DH Let BC, CG be placed in a straight line ; therefore DC and CE are also in a straight line (14.
Side 2 - A rhomboid, is that which has its opposite sides equal to one another, but all its sides are not equal, nor its angles right angles.
Side 105 - The first of four magnitudes is said to have the same ratio to the second, which the third has to the fourth, when any equimultiples whatsoever of the first and third being taken, and any equimultiples whatsoever of the second and fourth ; if the multiple of the first be less than that of the second, the multiple of the third is also less than that of the fourth...
Side 79 - 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.
Side 1 - A straight line is that which lies evenly between its extreme points.
Side 149 - 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 - That, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.
Side 83 - Wherefore from the given circle ABC has been cut off the segment BAC, containing an angle equal to the given angle DQEP PROP. XXXV. THEOR. If two straight lines within a circle cut one another, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other. Let the...