The Synoptical Euclid; Being the First Four Books of Euclid's Elements of Geometry, from the Edition of Dr. Robert Simson ... With ExercisesCharles Henry Law, 1854 - 120 sider |
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Resultat 1-5 av 38
Side 7
... point C , in which the circles cut one another , draw ( Post . 1. ) the straight lines CA , CB , to the points 4 , B ... F. From the centre B at the distance BC , describe ( Post . 3. ) the circle CGH , and from the centre D , at ...
... point C , in which the circles cut one another , draw ( Post . 1. ) the straight lines CA , CB , to the points 4 , B ... F. From the centre B at the distance BC , describe ( Post . 3. ) the circle CGH , and from the centre D , at ...
Side 9
... point C shall coincide with the point F , because the straight line AC is equal to DF . But the point B coincides with the point Ę ; wherefore the base BC shall coincide with the base EF ; because , the point B coinciding with E , and C ...
... point C shall coincide with the point F , because the straight line AC is equal to DF . But the point B coincides with the point Ę ; wherefore the base BC shall coincide with the base EF ; because , the point B coinciding with E , and C ...
Side 10
... point F , and from AE , the greater , cut off AG equal ( I. 3. ) to AF , the less , and join FC , GB . Because AF is equal ( Constr . ) to 4G , and AB ( Hypothesis ) to AC , the two sides FA , AC are equal to the two GA , AB , each to ...
... point F , and from AE , the greater , cut off AG equal ( I. 3. ) to AF , the less , and join FC , GB . Because AF is equal ( Constr . ) to 4G , and AB ( Hypothesis ) to AC , the two sides FA , AC are equal to the two GA , AB , each to ...
Side 13
... point B be on E , and the straight line BC upon EF ; 1 . The point C shall coincide with the point F , because BC is equal to EF ; therefore BC coinciding with EF , 2 . BA and AC shall coincide with ED and DF ; for , if the base BC ...
... point B be on E , and the straight line BC upon EF ; 1 . The point C shall coincide with the point F , because BC is equal to EF ; therefore BC coinciding with EF , 2 . BA and AC shall coincide with ED and DF ; for , if the base BC ...
Side 16
... F , G ; and bisect ( I. 10. ) FG in H , and join CH ; the straight line CH , drawn from the given point C , is perpendicular to the given straight line AB . C E II G B D Join CF , CG ; and because FH is equal to HG , and HC common to ...
... F , G ; and bisect ( I. 10. ) FG in H , and join CH ; the straight line CH , drawn from the given point C , is perpendicular to the given straight line AB . C E II G B D Join CF , CG ; and because FH is equal to HG , and HC common to ...
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The Synoptical Euclid; Being the First Four Books of Euclid's Elements of ... Euclid,Samuel A. GOOD Uten tilgangsbegrensning - 1854 |
The synoptical Euclid; being the first four books of Euclid's Elements of ... Euclides Uten tilgangsbegrensning - 1853 |
Vanlige uttrykk og setninger
AB is equal AC is equal adjacent angles angle ABC angle ACB angle AGH angle BAC angle BCD angle EDF angle equal base BC bisected circle ABC circumference diameter double draw equal angles equal Constr equal Hyp equal straight lines equal to BC equilateral and equiangular EUCLID'S ELEMENTS exterior angle given circle given point given rectilineal angle given straight line given triangle gnomon greater inscribed interior and opposite less Let ABC Let the straight likewise opposite angles parallel to CD parallelogram pentagon perpendicular point F produced Q.E.D. PROP rectangle AE rectangle contained remaining angle required to describe right angles semicircle side BC square of AC straight line AB straight line AC straight line drawn touches the circle triangle ABC twice the rectangle