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18. A diagonal is a line which joins the vertices of two angles not adjacent, as AC (fig. 42).
Fig. 42 19. An equilateral polygon is one which has all its sides equal; an equiangular polygon is one which has all its angles equal.
20. Two polygons are equilateral with respect to each other, when they have their sides equal, each to each, and placed in the same order, that is, when by proceeding round in the same direction the first in the one is equal to the first in the other, the second in the one to the second in the other, and so on. similar sense are to be understood two polygons equiangular with respect to each other. The equal sides in the first case, and the equal angles in the second, are called homologous (A).
21. An Axiom is a proposition, the truth of which is self-evident.
A Theorem is a truth which becomes evident hy a process of reasoning called a demonstration.
A Problem is a question proposed which requires a solution.
A Lemma is a subsidiary truth employed in the demonstration of a theorem, or in the solution of a problem.
The common name of Proposition is given indifferently to theorems, problems, and lemmas.
A Corollary is a consequence which follows from one or several propositions.
A Scholium is a remark upon one or more propositions which have gone before, tending to show their connexion, their restriction, their extension, or the manner of their application.
A Hypothesis is a supposition made either in the enunciation of a proposition, or in the course of a demonstration.
22. Two quantities, each of which is equal to a third, are equal to one another.
23. The whole is greater than its part.
26. Two magnitudes, whether they be lines, surfaces, or solids, are equal, when, being applied the one to the other, they coincide with each other entirely, that is, when they exactly fill the same space.
OF PLANE FIGURES.
27. All right angles are equal.
Demonstration. Let the straight line CD be perpendicular to Fig. 16. AB (fig. 16), and GH to EF, the angles ACD, EGH, will be
Take the four distances CA, CB, GE, GF, equal to each other,
KCB< BCD; besides, by hypothesis,
ACD = BCD; hence
ACK>KCB. and the line GH cannot fall upon a line CK different from CD; consequently it falls upon CD, and the angle EGH upon ACD, and EGH is equal to ACD; therefore all right angles are equal.
28. A straight line CD (fig. 17), which meets another straight line AB, makes with it two adjacent angles ACD, BCD, which, taken together, are equal to two right angles.
Demonstration. At the point C, let CE be perpendicular to AB. The angle ACD is the sum of the angles ACE, ECD; therefore ACD + BCD is the sum of the three angles ACE, ECD, BCD. The first of these is a right angle, and the two others are together equal to a right angle; therefore the sum of the two angles ACD, BCD, is equal to two right angles.
29. Corollary I. If one of the angles ACD, BCD, is a right angle, the other is also a right angle.
30. Corollary 11. If the line DE (fig. 18) is perpendicular to Fig. 18. AB; reciprocally, AB is also perpendicular to DE.
For, since DE is perpendicular to AB, it follows that the angle ACD is equal to its adjacent angle DCB, and that they are both right angles. But, since the angle ACD is a right angle, it follows that its adjacent angle ACE is also a right angle; therefore the angle ACE= ACD, and AB is perpendicular to DE.
31. Corollary m. All the successive angles, BAC, CAD, DAE, EAF, (fig. 34), formed on the same side of the straight Fig. 34. line BF, are together equal to two right angles; for their sum is equal to that of the two angles BAM, MAF; AM being perpendicular to BF.
32. Two straight lines, which have two points common, coincide throughout, and form one and the same straight line.
Demonstration. Let the two points, which are common to the two lines, be A and B (fig. 19). In the first place it is evident Fig. 19. that they must coincide entirely between A and B ; otherwise, two straight lines could be drawn from A to B, which is impossible (25). Now let us suppose, if it be possible, that the lines, when produced, separate from each other at a point C, the one becoming CD, and the other CE. At the point C, let CF he drawn, so as to make the angle ACF, a right angle; then, ACD being a straight line, the angle FCD is a right angle (29); and, because ACE is a straight line, the angle FCE is a right angle. But the part FCE cannot be equal to the whole FCD; whence . straight lines, which have two points common A and B, cannot separate the one from the other, when produced; therefore they must form one and the same straight line.
33. If two adjacent angles ACD, DCB (fig. 20), are together equal to two right angles, the iwo exterior sides AC, CB, are in the same straight line.
Demonstration. For if CB is not the line AC produced, let CE be that line produced; then, ACE being a straight line, the angles ACD, DCE, are together equal to two right angles (28); but, by hypothesis, the angles ACD, DCB, are together equal to two right angles, therefore ACD+DCB = ACD + DCE. Take away the common angle ACD, and there will remain the part DCB equal to the whole DCE, which is impossible; therefore CB is the line AC produced.
34. Whenever two straight lines AB, DE (fig. 21), cut each other, the angles oppositet to each other at the vertex are equal.
Demonstration. Since DE is a straight line, the sum of the angles ACD, ACE, is equal to two right angles; and, since AB is a straight line, the sum of the angles ACE, BCE, is equal to two right angles; therefore ACD + ACE=ACE + BCE; from cach of these take away the common angle ACE, and there will remain the angle ACD equal to its opposite angle BCE.
It may be demonstrated, in like manner, that the angle ACE is equal to its opposite angle BCD.
35. Scholium. The four angles, formed about a point by two straight lines which cut each other, are together equal to four right angles; for the angles ACE, BCE, taken together, are equal to two right angles; also the other angles ACD, BCD, are together equal to two right angles.
In general, if any number of straight lines, as CA, CB (fig. 22), &c., meet in the same point C, the sum of all the successive angles, ACB, BCD, DCE, ECF, FCA, will be equal to four right angles. For, if at the point C, four right angles be formed by two lines perpendicular to each other, they will comprehend the same space as the successive angles, ACB, BCD, &c.
† These are often called vertical angles.
36. Two triangles are equal, when two sides and the included angle of the one are equal to two sides and the included angle of the other, each to each.
Demonstration. In the two triangles ABC, DEF (fig. 23), let Fig. 23. the angle A be equal to the angle D, the side AB equal to the side DE, and the side AC equal to the side DF; the two triangles ABC, DEF, will be equal.
Indeed the triangles may be so placed, the one upon the other, that they shall coincide throughout. If, in the first place, we apply the side DE to its equal AB, the point D will fall upon A, and the point E upon B. But, since the angle D is equal to the angle A, when the side DE is placed apon AB, the side DF will take the direction AC; moreover DF is equal to AC; therefore the point F will fall upon C, and the third side EF will exactly coincide with the third side BC; therefore the triangle DEF is. equal to the triangle ABC (26).
37. Corollary. When, in two triangles, these three things are equal, namely, the angle A= D, the side AB = DE, and the side
DF, we may thence infer, that the other three are also equal, namely, the angle B= E, the angle C = F, and the side BC = EF.
7 38. Two triangles are equal, when a side and the two adjacent angles of the one, are equal to a side and the two adjacent angles of the other, each to each.
Demonstration. Let the side BC (fig. 23) be equal to the side Fig. 23. EF, the angle B equal to the angle E, and the angle C equal to the angle F; the triangle ABC will be equal to the triangle DEF.
For, in order to apply the one to the other, let EF be placed upon its equal BC, the point E will fall upon B and the point F upon C. Then because the angle E is equal to the angle B, the side ED will take the direction BA, and therefore the point D will be somewhere in BA ; also because the angle F is equal to C, the side FD will take the direction CA, and therefore the point D will be somewhere in CA ; whence the point D, which must be at the same time in the lines BA and CA, can only be at their intersection A ; therefore the two triangles ABC,