The first three books of Euclid's Elements of geometry, with theorems and problems, by T. Tate1851 - 139 sider |
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Side 56
... construct the figure as in the preceding propo- sitions : And because AG is equal ( 1 43. ) to GE , add to each of them CK ; the whole AK is therefore equal to the whole CE ; therefore AK , CE are double of AK ; but AK , CE are the H ...
... construct the figure as in the preceding propo- sitions : And because AG is equal ( 1 43. ) to GE , add to each of them CK ; the whole AK is therefore equal to the whole CE ; therefore AK , CE are double of AK ; but AK , CE are the H ...
Side 57
... construct two figures such as in the preceding . Because CB is equal to BD , and that CB is equal ( 1. 34. ) to GK , and BD to KN ; therefore GK is equal to KN : For the same reason , PR is equal to RO ; and because CB is equal to BD ...
... construct two figures such as in the preceding . Because CB is equal to BD , and that CB is equal ( 1. 34. ) to GK , and BD to KN ; therefore GK is equal to KN : For the same reason , PR is equal to RO ; and because CB is equal to BD ...
Side 110
... construct it . 8. Given the base , the perpendicular height , and one of the angles at the base of a triangle , to ... construct the triangle . 13. To construct a right - angled triangle , having 110 GEOMETRICAL EXERCISES .
... construct it . 8. Given the base , the perpendicular height , and one of the angles at the base of a triangle , to ... construct the triangle . 13. To construct a right - angled triangle , having 110 GEOMETRICAL EXERCISES .
Side 111
... construct it . Show that , in general , there are two triangles answering the conditions of the problem . 15. Given the base , one of the other sides , and the perpendicular on the base of a triangle , to con- struct it . 16. On a given ...
... construct it . Show that , in general , there are two triangles answering the conditions of the problem . 15. Given the base , one of the other sides , and the perpendicular on the base of a triangle , to con- struct it . 16. On a given ...
Side 115
... constructed upon the two diagonals of the rhombus . 29. Parallelograms whose sides and angles are equal are themselves equal . 30. The lines which join the middle points of the three sides of a triangle , divide it into four equal ...
... constructed upon the two diagonals of the rhombus . 29. Parallelograms whose sides and angles are equal are themselves equal . 30. The lines which join the middle points of the three sides of a triangle , divide it into four equal ...
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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