3. The whole is equal to all its parts. 4. If to equal things, equal things be added, the wholes are equal. 5. If from equal things, equal things be deducted, the remainders are equal. 6. If to or from unequal things, equal things be added or taken, the sums or remainders are 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. NOTES. 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 A scholium, is a remark or observation made upon something going before. SIGNIFICATION OF SIGNS. The sign, denotes the quantities between which it stands to be equal. The sign+, denotes the quantity it precedes to be added. The sign, denotes the quantity which it precedes to be subtracted. The sign x, denotes the quantities, between them to be multiplied into each other. To denote that four quantities, A, B, C, D, are proportional, they are usually written thus, A:B:: C: D; and read thus, as A is to B, so is C to D; but when three quantities A, B, C, are proportional, the middle quantity is repeated, and they are written A:B::B: C. GEOMETRICAL THEOREMS. THEOREM I. IF 1. If AB be perpendicular to CD, then (by def. 11.) 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 to two right angles. Q. E. D. Corollary 1. Whence if any number 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 angles, 2. And all the angles which can be made about a THEOREM II. If one right line cross another (as C does BD) the opposite angles ade by those lines, will be equal to ach other; that is, AEB to CED nd BEC to AED. fig. 21. 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 5.) Q. E. D. After the same manner CED+AED=2_right angles; and AED+AEB=2 right angles; wherefore taking AED from both, there remains CED-AEB. Q. E. D. 4. Their external angles are equal to each other, that is, GEB-CFH. 2. The alternate angles will be equal, that is, AEF =EFD and BEF-CFE. 3. The external angle will be equal to the internal and opposite 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 two 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 CFH (by part 1. of this theo.) therefore AEF= EFD. The same way we prove FEC=EFB. 3. AEF=EFD; (by the last part of this theo.) but AEF 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 GEB+ 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 ere equal |