The text of Euclid's geometry, book 1, uniformly and systematically arranged by J.D. Paul1884 |
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Page 72
... Constr . Def . 15 . Because B is the centre of the circle ACE ; Constr . therefore BC is equal to AB . Because AC and BC are each equal to AB ; therefore AC is equal to BC . Because AC , AB , and BC are equal to one another ; therefore ...
... Constr . Def . 15 . Because B is the centre of the circle ACE ; Constr . therefore BC is equal to AB . Because AC and BC are each equal to AB ; therefore AC is equal to BC . Because AC , AB , and BC are equal to one another ; therefore ...
Page 74
... Constr . Def . 15 . Constr . Def . 15 . Proved . Constr . Axiom . 3 . Proved . Axiom . 1 . Therefore , from the given point A , a straight line AL has been drawn equal to the given straight line BC . Q. E. F. Postulate 1. A straight ...
... Constr . Def . 15 . Constr . Def . 15 . Proved . Constr . Axiom . 3 . Proved . Axiom . 1 . Therefore , from the given point A , a straight line AL has been drawn equal to the given straight line BC . Q. E. F. Postulate 1. A straight ...
Page 76
... Constr . Def . 15 . Proved . Constr . Axiom 1 . Therefore , from AB , the greater of two given straight lines , a part AE has been cut off , equal to CG the less . Q. E. F. Proposition II . A straight line may be drawn from 76 GEOMETRY .
... Constr . Def . 15 . Proved . Constr . Axiom 1 . Therefore , from AB , the greater of two given straight lines , a part AE has been cut off , equal to CG the less . Q. E. F. Proposition II . A straight line may be drawn from 76 GEOMETRY .
Page 80
... Constr . Нур . g . I. 4 . Because AF is equal to AG , and AB is equal to AC ; Constr . Нур . therefore BF is equal to CG . Axiom 3 . Because BF is equal to CG , In the triangles BFC and CGB ; and FC is and the angle BFC is therefore the ...
... Constr . Нур . g . I. 4 . Because AF is equal to AG , and AB is equal to AC ; Constr . Нур . therefore BF is equal to CG . Axiom 3 . Because BF is equal to CG , In the triangles BFC and CGB ; and FC is and the angle BFC is therefore the ...
Page 82
... Therefore if two angles of a triangle & c . Constr . g . Hyp . I. 4 . Proved . Axiom 9 . G. Q. E. D. Corollary . Hence every equiangular triangle is also equilateral . Proposition III . From the greater of two given straight 82 GEOMETRY .
... Therefore if two angles of a triangle & c . Constr . g . Hyp . I. 4 . Proved . Axiom 9 . G. Q. E. D. Corollary . Hence every equiangular triangle is also equilateral . Proposition III . From the greater of two given straight 82 GEOMETRY .
Common terms and phrases
2nd Edition 4th Edition AB is equal ABC and DEF ABC is equal adjacent angles angle ABC angle ACB angle AGH angle BAC angle BCD angle CAB angle contained angle DEF angle EDF angle equal angle GHD angles are equal angles CBA Axiom 9 base BC BDC is greater bisector bisects centre circle coincide Constr Crown 8vo describe diagonal equal sides equal to BD equal to EF equilateral triangle Euclid exterior angle F. A. Paley Fcap finite straight line Geometry given point given straight line isosceles triangle Join Latin magnitude are equal opposite sides parallel to BC parallelogram parallelogram ABCD plane figure Post 8vo Postulate produced Q. E. D. PROPOSITION quadrilateral rectilineal angle REQUIRED TO PROVE rhombus right angles sides equal square on AC straight line drawn THEOREM triangle ABC triangle DEF triangles are equal vertex Нур
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Page 72 - LET it be granted that a straight line may be drawn from any one point to any other point.
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