A companion to Euclid: being a help to the understanding and remembering of the first four books. With a set of improved figures, and an original demonstration of the proposition called in Euclid the twelfth axiom, by a graduateJohn W. Parker, 1837 - 88 sider |
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Side 19
... greater , which shows the supposition to be false , 4. that similarly * no other line but BD is in the same right line with BC . PROPOSITION XV . Theorem . If two straight lines cut one another , the vertical , or opposite , angles ...
... greater , which shows the supposition to be false , 4. that similarly * no other line but BD is in the same right line with BC . PROPOSITION XV . Theorem . If two straight lines cut one another , the vertical , or opposite , angles ...
Side 20
... greater than either of the interior opposite angles . B B Steps of the Demonstration . Jbase AB = base FC , 1. Prove ... greater side of every triangle is opposite to the greater angle . B Steps of the Demonstration . 1. Prove that ABD ...
... greater than either of the interior opposite angles . B B Steps of the Demonstration . Jbase AB = base FC , 1. Prove ... greater side of every triangle is opposite to the greater angle . B Steps of the Demonstration . 1. Prove that ABD ...
Side 21
... greater angle of every triangle is subtended by the greater side , or has the greater side opposite to it . B B Steps of the Demonstration . 1. Prove that AC cannot be AB , that AC cannot be < AB , 2 . 3 . that . Ac must be > AB ...
... greater angle of every triangle is subtended by the greater side , or has the greater side opposite to it . B B Steps of the Demonstration . 1. Prove that AC cannot be AB , that AC cannot be < AB , 2 . 3 . that . Ac must be > AB ...
Side 22
... greater angle . Steps of the Demonstration . 1. Prove that BA + AC > BE + EC , that BE + EC > CD + DB , that much more is BA + AC > CD + Db , 2 . 3 . 4 . that BDC > BAC . D PROPOSITION XXII . A- B Problem . To make a triangle , the ...
... greater angle . Steps of the Demonstration . 1. Prove that BA + AC > BE + EC , that BE + EC > CD + DB , that much more is BA + AC > CD + Db , 2 . 3 . 4 . that BDC > BAC . D PROPOSITION XXII . A- B Problem . To make a triangle , the ...
Side 23
... greater angle shall G be greater than the base of the other . B F Steps of the Demonstration . 1. Prove that ( in As ABC , DEG ) base BC = base EG , that DFG DGF , 2 . 3 . that DFG > EGF , 4 . that much more EFG > EGF , 5 . that . EG or ...
... greater angle shall G be greater than the base of the other . B F Steps of the Demonstration . 1. Prove that ( in As ABC , DEG ) base BC = base EG , that DFG DGF , 2 . 3 . that DFG > EGF , 4 . that much more EFG > EGF , 5 . that . EG or ...
Vanlige uttrykk og setninger
AB² AC² AD² AEX EC angle contained angle equal Argument ad absurdum base DF BC² BD² bisect CB² cuts the circle DC² Demonstration itself consists diameter EB² EF² EG² Engravings equal straight lines equi equiangular equilateral Euclid F Steps fall figure GF² given circle given point given rectilineal angle given straight line given triangle i. e. less inscribe interior angles learner less greater line be divided line drawn parallel parallelogram PARKER pass pentagon point of contact Problem proof PROPOSITION IX PROPOSITION VIII Proved by showing rectangle contained right angles right line shows the supposition similarly Suppose supposition is false Theorem WEST STRAND whole line
Populære avsnitt
Side 24 - If two triangles have two angles of the [one equal to two angles of the other, each to each, and one side equal to one side, namely, either t}le sides adjacent to the equal...
Side 45 - To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts, shall be equal to the square of the other part.
Side 18 - If, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line.
Side 61 - From this it is manifest that the straight line which is drawn at right angles to the diameter of a circle from the extremity of it, touches the circle...
Side 37 - In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.
Side 76 - IF from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it ; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square of the line which touches it.
Side 77 - If from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it, and if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, be equal to the square on GEOMETRY.
Side 72 - If a straight line touch a circle, and from the point of contact a straight line be drawn at right angles to the touching line, the centre of the circle shall be in that line.
Side 27 - If a straight line fall on two parallel straight lines, it makes the alternate angles equal to one another, and the exterior angle equal to the interior and opposite angle on the same side; and also the two interior angles on the same side together equal to two right angles.