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. 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. Ą 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. GEOMETRICAL THEOREMS, THEOREM I. PL. 1, fig. 20. IF a right line falls on another, as AB, or EB, does on CD, it either makes with it two 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. 2. 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. THEO. II. P PL. 1. fig. 21. If one right line cross another, (as AC does BD) the opposite angles made by those lines, will be equal 10 each other : that is, AEB to CED and BEC to AED. By theorem 1. BEC + CED and CED + DEA 2 right angles. 2 right angles. Therefore (by axiom 1.) BEC+CED=CED+ DEA: take CED from both, and there remains BEC = DEA. (by axiom 5.) 2. E. D. After the same manner CED + AED= 2 right angles; and AED+AEB=tworight angles; wherefore taking AED from both, there remains CED = AEB. 2. E. D. AEO. III. Pl. 1. fig. 22. If a right line cross two parallels, as GH does AB and CD, then, 1. Their external angles are equal to each other, that zs, 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 tivo 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 FEB = EFC. 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 AEFECFE are equal to two right angles. 2. E. D. THEO. IV. PL. 1. 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 two internal opposite angles taken together. Viz. to B and A. Through Colet 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 cats the same parallels, the angle ACE = A (by part 2. of the last.) Therefore ECD+ACE= ACD=B+A. 2. E. D. THEO. V. PL. l. 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 H Н —the given angle will be equal to the sum of the other two; or 180—the sum of two given angles, gives the other one. Cor. 2. In every right-angled triangle, the two acute angles are=90 degrees, or to one right angle: therefore 90-one acute angle, gives the other. THEO. VI. Pl. l. fig. 24. If in any two triangles, ABC, DEF, there be two sides, AB, AC in the one, severally equal to DE, DF in the other, and the angle A contained between the two sides in the one, equal to D in the other; then the remaining angles of the one, will be severally equal to those of the other, viz. B=E and C= F: and the base of the one BC, will be equal to EF, that of the other. If the triangle ABC be supposed to be laid on the triangle DEF, so as to make the points A and B coincide with Dand E, which they will do, because AB=DE (by the hypothesis); and since the angle A=D, the line AC will fall along DF, and inasmuch as they are supposed equal, C will fall in F; seeing therefore the three points of one coincide with those of the other triangle, they are manifestly ecual to each other ; therefore the angle B=E and C=F, and BC=EF. 2. E. D. LEMMA. PL. 1. fig. 11. If two sides of a triangle a b cbe equal to each other, that is, ac=cb the angles which are opposite to those equal sides, will also be equal to each other ; viz, a=b. For let the triangle a b c be divided into two |