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 σελίδες |
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Αποτελέσματα 1 - 5 από τα 72.
Σελίδα 10
... join FC , GB . Because AF is equal ( Constr . ) to AG , and AB ( Hypothesis ) to AC , the two sides FA , AC are equal to the two GA , AB , each to each ; and they contain the angle FAG common to the two triangles AFC , AGB ; therefore ...
... join FC , GB . Because AF is equal ( Constr . ) to AG , and AB ( Hypothesis ) to AC , the two sides FA , AC are equal to the two GA , AB , each to each ; and they contain the angle FAG common to the two triangles AFC , AGB ; therefore ...
Σελίδα 11
... join DC . Therefore because in the triangles DBC , ACB , DB is equal to AC , and BC common to both , 1. The two sides DB , BC , are equal to the two AC , CB , each to each ; and the angle DBC is equal ( Hyp . ) to the angle ACB ...
... join DC . Therefore because in the triangles DBC , ACB , DB is equal to AC , and BC common to both , 1. The two sides DB , BC , are equal to the two AC , CB , each to each ; and the angle DBC is equal ( Hyp . ) to the angle ACB ...
Σελίδα 12
... Join CD . Then , in the case in which the vertex of each of the tri- angles is without the other triangle , because AC is equal to AD , ( I. 5. ) 1. The angle ACD is equal to the angle ADC . But the angle ACD is greater than the angle ...
... Join CD . Then , in the case in which the vertex of each of the tri- angles is without the other triangle , because AC is equal to AD , ( I. 5. ) 1. The angle ACD is equal to the angle ADC . But the angle ACD is greater than the angle ...
Σελίδα 13
... join DE , and upon it , on the side remote from 4 , describe ( I. 1. ) an equilateral triangle DEF ; then join AF ; the straight line AF bisects the triangle BAC . A B Because AD is equal to AE , and AF is common to the two triangles ...
... join DE , and upon it , on the side remote from 4 , describe ( I. 1. ) an equilateral triangle DEF ; then join AF ; the straight line AF bisects the triangle BAC . A B Because AD is equal to AE , and AF is common to the two triangles ...
Σελίδα 14
... angles to AB . Take any point D in AC , and ( I. 3. ) make CE equal to CD , and upon DE describe ( I. 1. ) the equilateral triangle DFE , and join FC ; the straight line FC drawn from the given point is at 14 EUCLID'S ELEMENTS .
... angles to AB . Take any point D in AC , and ( I. 3. ) make CE equal to CD , and upon DE describe ( I. 1. ) the equilateral triangle DFE , and join FC ; the straight line FC drawn from the given point is at 14 EUCLID'S ELEMENTS .
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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 THEOREM touches the circle triangle ABC twice the rectangle