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 22.
Σελίδα 9
... impossible ; therefore 4. The base BC shall coincide with the base EF , and ( Ax . 8. ) be equal to it . Wherefore also 5. The whole triangle ABC shall coincide with the whole triangle DEF and be equal to it ; and the other angles of ...
... impossible ; therefore 4. The base BC shall coincide with the base EF , and ( Ax . 8. ) be equal to it . Wherefore also 5. The whole triangle ABC shall coincide with the whole triangle DEF and be equal to it ; and the other angles of ...
Σελίδα 12
... impossible . But if one of the vertices , as D , be within the other triangle ACB ; produce AC , AD to E , F ; therefore because AC is equal to AD in the triangle ACD , ( I. 5. ) the angles upon the other side of the base CD , namely ...
... impossible . But if one of the vertices , as D , be within the other triangle ACB ; produce AC , AD to E , F ; therefore because AC is equal to AD in the triangle ACD , ( I. 5. ) the angles upon the other side of the base CD , namely ...
Σελίδα 13
... impossible ( I. 7. ) ; therefore , if the base BC coincide with the base EF , the sides BA , AC , cannot but coincide with the sides ED , DF ; wherefore likewise 3. The angle BAC coincides with the angle EDF , and is equal ( Ax . 8 ...
... impossible ( I. 7. ) ; therefore , if the base BC coincide with the base EF , the sides BA , AC , cannot but coincide with the sides ED , DF ; wherefore likewise 3. The angle BAC coincides with the angle EDF , and is equal ( Ax . 8 ...
Σελίδα 15
... 3. The angle DBE is equal to the angle CBE , the less to the greater , which is impossible ; therefore 4 . Two straight lines cannot have a common segment . PROP . XII . - PROBLEM . To draw a BOOK I. - PROP . XI . 15.
... 3. The angle DBE is equal to the angle CBE , the less to the greater , which is impossible ; therefore 4 . Two straight lines cannot have a common segment . PROP . XII . - PROBLEM . To draw a BOOK I. - PROP . XI . 15.
Σελίδα 18
... impossible ; therefore 4. BE is not in the same straight line with BC . And in like manner it may be demonstrated that no other can be in the same straight line with it but BD , therefore 5. BD is in the same straight line with CB ...
... impossible ; therefore 4. BE is not in the same straight line with BC . And in like manner it may be demonstrated that no other can be in the same straight line with it but BD , therefore 5. BD is in the same straight line with CB ...
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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 touches the circle triangle ABC twice the rectangle