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POSTULATES OR PETITIONS.
1. That a right line may be drawn from any one given point to another.
2. That a right line may be produced or continued at pleasure.
3. That from any centre and with any radius, the circumference of a circle may be described.
4. It is also required that the equality of lines and angles to others given, be granted as possible : that it is possible for one right line to be perpendicularto another, at a given point or distance; and that every magnitude has its half, third, fourth,
Note, Though these postulates are not always quoted, the reader will easily perceive where, and in what sense they are to be understood.
AXIOMS or self-evidlent TRUTHS.
1. Things that are equal to one and the same thing, are equal to each other.
2. Every whole is greater than its part.
3. Every whole is equal to all its parts taken together.
4. If to equal things, equal things be added, the whole will be equal.
5. If from equal things, equal things he deducted, the remainders will be equal.
6. If to or from unequal things, equal things be added or taken, the sums or remainders will be unequal.
7. All right angles are equal to one another.
8. If two right lines not parallel, be produced towards their nearest distance, they will intersect each other.
9. Things which mutually agree with each other, are equal.
A theorem is a proposition, wherein something is proposed to be demonstrated.
A problem is a proposition, wherein something is to be done or effected.
A lemma is some demonstration, previous and necessary, to render what follows the more easy.
A corollary is a consequent truth, deduced from a foregoing demonstration.
A scholium, is a remark or observation made upon something going before.
PL. 1, fig. 20.
If a right line falls on another, as AB, or EB, does on CD, it either makes with it tevo right angles, or two angles equal to two right angles.
1. If AB be perpendicular to CD, then (by def. 10.) the angles CBA, and ABD, will be each a right angle.
2. But if EB fall slantwise on CD, then are the angles DBE+EBC=DBE+EBA(=DBA)+ ABC, or two right angles. Q. E. D.
Corollary 1. Whence if any numbers of right lines were drawn from one point, on the same side of a right line; all the angles made by these lines will be equal to two right lines.
2. And all the angles which can be made about a point, will be equal to four right angles.
PL. l. fig. 21.
If one right line cross another, (as AC does BD) the oppo. site angles made by those lines, will be equal to each other : that is, AEB to CED, and BEC to AED.
By theorem 1. BEC + CED = 2 right angles.
and CED + DEA= 2 right angles. Therefore (by axiom 1.) BEC+CED=CED+ DEA: take CED from both, and there remains BEC=DEA. (by axiom 3.) Q. E. D.
After the same manner CED+AED=2 right augles; and AED + AEB = two right angles ; wherefore taking AED from both, there remains CED=AEB. Q. E. D.
Pl. l. fig. 22. If a right line cross tro parallels, as GH docs AB and CD, then,
1. Their external angles are equal to each other, that is, GEB = CFH.
2. The alternate angles will be equal, that is, 1EP EFD and BEF CFE.
3. The external angle will be equal to the internal and opposile one on the same side, that is, GEB EFD and AEG CFE.
4. And the sum of the internal angles on the same side, are equal to two right angles ; that is, BEF+ DFE are equal to two right angles, and AEF + CFE are equal to 1720 right angles.
1. Since AB is parallel to CD, they may be considered as one broad line, crossed by another line, as GH;(then by the last theo.) GEB=CFH, and AEG=HFD.
2. Also GEB= AEF, and CFH= EFD; but GEB= CFI (by part 1. of this theo.) therefore AEF=EFD. The same way we prove FEB= EFC.
3. AEF=EFD; (by the last part of this theo. but 4EF=GEB (by theo. 2.) Therefore GEB -EFD. The same way we prove AEG =CFE. 4. For since GEB= EFD, to both add FEB, then (by axiom 4.)GEB +FEB=EFD +FEB, but ĜĖB + FEB, are equal to two right angles (by theo. 1.) Therefore EFD + FEB are equal to two right angles : after the same manner we prove that AEF + CFE are equal to two right angles. Q. E. D.
Pl. l. fig. 23.
In any triangle ABC, one of its legs, as BC, being produced towards D, it will make the external angle ACD equal to the ito internal opposite angles taken together. Viz. to B and A.
Through C, let CE be drawn parallel to AB; then since BD cuts the two parallel lines BA, CE; the angle ECD = B, (by part 3. of the last theo.) and again, since AC cuts the same parallels, the angle ACE = A (by part. 2. of the last.) Therefore ECD + ACÉE ACD =B + A. Q. E. D.
Pl. 1. fig. 23. In any triangle ABC, all the three angles, taken together, are equal to two right angles, viz. A + B + ACB = 2 right angles.
Produce CB to any distance, as D, then (by the last) ACD=B+A; to both add ACB; then ACD + ACB= A + B + ACB; but ACD + ACB =2 right angles (by theo. 1.); therefore the three angles A + B + ACB = 2 right angles. Q. E. D.
Cor. 1. Hence if one angle of a triangle be known, the sum of the other two is also known : for since the three angles of every triangle contain two right ones, or 180 degrees, therefore 180