The first three books of Euclid's Elements of geometry, with theorems and problems, by T. Tate1851 - 139 sider |
Inni boken
Resultat 6-10 av 36
Side 37
... bisects them ; for AB being equal to CD , and BC common , the two A B , BC are equal to the two DC , CB , each to each ; and the angle ABC is equal to the angle BCD ; therefore the triangle ABC is equal ( 1. 4. ) to the triangle BCD ...
... bisects them ; for AB being equal to CD , and BC common , the two A B , BC are equal to the two DC , CB , each to each ; and the angle ABC is equal to the angle BCD ; therefore the triangle ABC is equal ( 1. 4. ) to the triangle BCD ...
Side 39
... bisects it ( 1. 34. ) ; and the triangle DBC is the half of the parallelogram DBCF , because the diameter DC bisects it : But the halves of equal things are equal ( Ax . 7. ) ; therefore the triangle ABC is equal to the triangle DBC ...
... bisects it ( 1. 34. ) ; and the triangle DBC is the half of the parallelogram DBCF , because the diameter DC bisects it : But the halves of equal things are equal ( Ax . 7. ) ; therefore the triangle ABC is equal to the triangle DBC ...
Side 40
... bisects it ; and the triangle DEF is the half ( 1. 34. ) of the parallelogram DEH , because the diameter DF bisects it : But the halves of equal things are equal ( Ax . 7. ) ; therefore the triangle ABC is equal to the triangle DEF ...
... bisects it ; and the triangle DEF is the half ( 1. 34. ) of the parallelogram DEH , because the diameter DF bisects it : But the halves of equal things are equal ( Ax . 7. ) ; therefore the triangle ABC is equal to the triangle DEF ...
Side 42
... BISECT ( 1. 10. ) B C in E , join A E , and at the point E in the straight line EC make ( 1.23 . ) the angle CEF equal to D ; and through a draw ( 1. 31. ) AG parallel to E C , and through c draw CG ( I. 31. ) paral- lel to E F ...
... BISECT ( 1. 10. ) B C in E , join A E , and at the point E in the straight line EC make ( 1.23 . ) the angle CEF equal to D ; and through a draw ( 1. 31. ) AG parallel to E C , and through c draw CG ( I. 31. ) paral- lel to E F ...
Side 62
... bisect ( 1. 10. ) AC in E , and join BE ; produce CA to F , and make ( 1. 3. ) EF equal to EB , and upon AF describe ( 1. 46. ) the square FGHA ; AB is divided in H , so that the rectangle AB , BH is equal to the square of AH . Produce ...
... bisect ( 1. 10. ) AC in E , and join BE ; produce CA to F , and make ( 1. 3. ) EF equal to EB , and upon AF describe ( 1. 46. ) the square FGHA ; AB is divided in H , so that the rectangle AB , BH is equal to the square of AH . Produce ...
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Vanlige uttrykk og setninger
A B C ABCD adjacent angles angle ABC angle ACB angle BAC angle equal angles BGH base BC BC is equal bisect centre circle ABC circumference diameter divided double draw a straight equal angles equal circles equal straight lines equal to FB exterior angle given point given rectilineal angle given straight line gnomon greater half a right hypotenuse interior and opposite isosceles triangle less Let ABC Let the straight line be drawn opposite angles parallel parallelogram perpendicular PROB produced Q. E. D. PROP rectangle AE rectangle contained rectilineal figure remaining angle right angles segment semicircle side BC square of AC straight line AB straight line AC straight line drawn THEOR touches the circle trapezium triangle ABC twice the rectangle vertex vertical angle