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E. 1, Hyp. At. B in AB, let BC and
BD make the adjacent
= two rt. angles;
one and the same st.
line CD C. ll by Sup. Should BD not be if the
same st. line with CB, 2 Pst. 2. make BE in the same st. line with CB. D. 11 by Hyp. :: AB with the st. line CBE makes the Zs ABC, ABE,
| But Zs ABC and ABD also = two rt. angles,
impossible; Conc. 1.:. BE is not in the same st. line with CB. Sim. In the same way no line except BD is in the same st. line
with CB; 10 Conc. 1... BD is in one and the same st. line with CB. 11! Rec. Wherefore if at a point in a line, &c.
Q.E.D. ScH.-The words “upon the opposite sides of it" are of essential importance; for two lines may make with a third, two angles equal to two rt. angles, and yet the two lines not be in one st. line.
Cor. 2.- And all the angles formed by any number of st. lines, AC, BC, DC, EC, FC, 8c., diverging from a common centre, C, are together equal to four ri. angles, C. l by Pst. 1. Produce any two of the
given lines, as AC to G,
BC to H.
in the point C,
ACH and HCG toge
ther = four rt. Zs.
HCG; 4 Con. .. all the angles diverging from · C are together equal to
I four rt. angles. Sch.-1. This proposition might be briefly proved by saying that the opposite angles are equal, because they have the same supplement.
2. “This Prop. is the developement of the definition of an angle. If the lines at the angular point be produced, the produced lines have the same inclination to one another as the original lines have.”-Potts' Euclid, p. 49.
3. The converse of this Prop. is, “ If foúr lines meeting in a point make the vertical angles equal, each alternate pair of lines shall be in one and the same st. line."
APP.-To find the breadth of a lake, or the distance between two inaccessible objects, d e and B.
From C, accessible both to A and B, measure the lines CA and CB, and produce them until CE is equal to CA, and CD to CB; join D and E, the distance DE will equal the distance AB.
For the lines CA, CB equal the lines CE, CD; and the angle ACB equals the angle DCE; therefore by Prop. 4, the lines AB and DE are equal.
Prop. 16.—THEOR. If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles.
Con.-P. 10, P. 3, Pst. 1, Pst. 2.-DEM.-P. 15, P. 4, A3. 9.
2 P. 16. But ext. ADB > int. _ DCB, 3 D. I, and ADB = LABD;
4 D. 2, 3. 1.. LABD > Z DCB; | 5 à fort. much more is L ABC > L ACB.
6 Rec. Therefore, the greater side of every triangle, &c. Q.E.D.
Sch.-The argument on which the conclusion depends is named “a fortiori," by the stronger reason, and proves that the given predicate belongs in a greater degree to one subject than to another; as in the Syllogism,-Y is greater than Z, and X greater than Y; much more is X greater than Z.
PROP. 19.-THEOR. (Converse of the 18th.) The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it.
DEM.-P. 5. P. 18.
SUP.-The side AC must be greater than, equal to, or less than AB. D. 1,64 Sup. & P. 5. , If AC = AB, then _ ABC = / ACB: 2 Hyp.
but _ABC + L ACB; 3 D. 1 & 2. 1.:. AC + AB. 41 Sup. & P. 18. Again, if AC is < AB, then _ ABC is< L ACB: 5 Hyp.
but ABC is * L ACB; 6 D. 4 & 5. 1.1. AC is * AB. 7 D. 3 & 6. | Now, AC is neither equal to, nor less than, AB;
1. AC is greater than AB. 9 Rec. Wherefore, the greater angle of every triangle, &c. Q.E.D.
SCH.-1. Props. 5, 6, 18, and 19, may be combined into one thus,-“ One angle of a triangle is greater than, equal to, or less than, another angle, as the side opposed to the one is greater than, equal to, or less than, the side opposed to the other; and vice versa.”
2. By Props. 17, 18, and 19, we may prove that from the same point there can be drawn but one perpendicular to a given line, and that this perpendicular is the shortest of all the lines from the given point to the given line.
3. As from a given point only three lines can be drawn perpendicular to each other, it is impossible to imagine that there are more than three species of quantity,-a Line a Surface, and a Solid.
Con.-Pst. 2, P. 3, Pst. 1.-DEM.-P. 5, Ax. 9, P. 19.
gether are > BC; sides AB and BC >'AC;
, BC and CA > AB. C. 1] by Pst. 2. Produce BA,
2 P. 3. and make AD = AC;
3 Pst. 1. join D and C. D. ilby C. 1 &P.5.1: AD= AC,
.: L ADC LACD;