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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Σελίδα 23
... reason , 2 . The exterior angle CEB of the triangle ABE is greater than BAC ; and it has been demonstrated that the angle BDC is greater than CEB ; much more then 3. The angle BDC is greater than the angle BAC . Therefore , if from the ...
... reason , 2 . The exterior angle CEB of the triangle ABE is greater than BAC ; and it has been demonstrated that the angle BDC is greater than CEB ; much more then 3. The angle BDC is greater than the angle BAC . Therefore , if from the ...
Σελίδα 36
... reason 5. The parallelogram EFGH is equal to the same EBCH . Therefore also 6. The parallelogram ABCD is equal to EFGH . Wherefore , parallelograms , & c . Q.E.D. PROP . XXXVII . - THEOREM . Triangles upon the same base and between the ...
... reason 5. The parallelogram EFGH is equal to the same EBCH . Therefore also 6. The parallelogram ABCD is equal to EFGH . Wherefore , parallelograms , & c . Q.E.D. PROP . XXXVII . - THEOREM . Triangles upon the same base and between the ...
Σελίδα 40
... reason , 3. The triangle KGC is equal to the triangle KFC . Then because the triangle AEK is equal to the triangle AHK , and the triangle KGC to KFC ; ( Ax . 2. ) 4. The triangle AEK together with the triangle KGC is equal to the ...
... reason , 3. The triangle KGC is equal to the triangle KFC . Then because the triangle AEK is equal to the triangle AHK , and the triangle KGC to KFC ; ( Ax . 2. ) 4. The triangle AEK together with the triangle KGC is equal to the ...
Σελίδα 44
... reason 2. AB and AH are in the same straight line . And because the angle DBC is equal to the angle FBA , each of them being a right angle , add to each the angle ABC , and ( Ax . 2. ) 3. The whole angle DBA is equal to the whole FBC ...
... reason 2. AB and AH are in the same straight line . And because the angle DBC is equal to the angle FBA , each of them being a right angle , add to each the angle ABC , and ( Ax . 2. ) 3. The whole angle DBA is equal to the whole FBC ...
Σελίδα 50
... reason , also , 10 . HF is a square , and it is upon the side HG , which is equal ( I. 34. ) to AC ; therefore 11 . HF , CK , are the squares of AC , CB . And because the complement AG is equal ( I. 43. ) to the complement GE , and that ...
... reason , also , 10 . HF is a square , and it is upon the side HG , which is equal ( I. 34. ) to AC ; therefore 11 . HF , CK , are the squares of AC , CB . And because the complement AG is equal ( I. 43. ) to the complement GE , and that ...
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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