The Synoptical Euclid; Being the First Four Books of Euclid's Elements of Geometry from the Edition of Dr. Robert Simson ... With Exercises. By S. A. Good ... Second Edition |
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Σελίδα 23
For the same 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 .
For the same 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 .
Σελίδα 35
to BC ; for the same reason , EF is equal to BC ; wherefore ( Ax . l . ) 1 . AD is equal to EF ; and DE is common ; therefore ( Ax . 2. or 3. ) 2 . The whole , or the remainder , AE is equal to the whole , or the remainder , DF ; AB ...
to BC ; for the same reason , EF is equal to BC ; wherefore ( Ax . l . ) 1 . AD is equal to EF ; and DE is common ; therefore ( Ax . 2. or 3. ) 2 . The whole , or the remainder , AE is equal to the whole , or the remainder , DF ; AB ...
Σελίδα 36
For the like 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 ...
For the like 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 ...
Σελίδα 40
The triangle AEK is equal to the triangle AHK ; and for the same 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 is equal to the triangle AHK ; and for the same 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 .
Σελίδα 44
For the same reason 2 . AB and AH are in the same straight line . And because the angle DBC is equal to the angle FB4 , each of them being a right angle , add to each the angle ABC , and ( Ax . 2. ) 3 . The whole angle DBA is equal to ...
For the same reason 2 . AB and AH are in the same straight line . And because the angle DBC is equal to the angle FB4 , each of them being a right angle , add to each the angle ABC , and ( Ax . 2. ) 3 . The whole angle DBA is equal to ...
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AB is equal ABC is equal ABCD angle ABC angle ACB angle AGH angle BAC angle BCD angle equal base BC bisected centre circle ABC circumference coincide common demonstrated describe diameter distance divided double draw equal angles exterior angle extremity fall figure four given circle given point given straight line gnomon greater impossible inscribed join less Let ABC Let the straight likewise manner meet opposite angles parallel parallelogram pass pentagon perpendicular point F PROBLEM produced Q.E.D. PROP reason rectangle contained rectilineal figure remaining angle right angles segment semicircle shown side BC sides square of AC straight line AC Take touches the circle triangle ABC twice the rectangle wherefore whole
Πληροφορίες βιβλιογραφίας
| Τίτλος | The Synoptical Euclid; Being the First Four Books of Euclid's Elements of Geometry from the Edition of Dr. Robert Simson ... With Exercises. By S. A. Good ... Second Edition |
| Συγγραφείς | EUCLID., Samuel A. GOOD |
| Έκδοση | 2 |
| Εκδότης | Charles Henry Law, 1854 |
| Πρωτότυπο από | Η Βρετανική βιβλιοθήκη |
| Ψηφιοποιήθηκε στις | 10 Δεκ. 2013 |
| Μέγεθος | 120 σελίδες |
| Εξαγωγή αναφοράς | BiBTeX EndNote RefMan |