are equal to fa, a g, each to each, and the base de to the base fg; the angle dce is equal (i. 8) to the angle fag. Therefore, at the given point a in the given straight line a b, the angle fag is made equal to the given rectilineal angle dce. Which was to be done. PROPOSITION XXIV.—THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of one of them greater than the angle contained by the two sides equal to them of the other; the base of that which has the greater angle shall be greater than the base of the other. LET a b c d e f be two triangles, which have the two sides a b, ac, equal to the two de, df, each to each, viz. a b equal to de, and a c to df; but the angle bac greater than the angle edf; the base bc is also greater than the base e f. Of the two sides de, d f, let de be the side which is not greater than the other, and at the point d, in the straight line de, make (i. 23) the angle e d.g. equal to the angle bac; and make dg equal (i. 3) to a c or df, and join eg; g f. Because a b is equal to d e, and a c to dg, the two sides ba, a c are equal to the two, ed, dg, each to each, and the angle bac is equal to the angle edg; there d fore the base bc is a equal (i. 4) to the base eg; and because dg is equal to df, the angle dfg is equal (i. 5) to the angle dgf; but the angle dgf is greater than the angle egf; there 8 fore the angle dfg b is greater than egf; f and much greater is the angle efg than the angle egf; and because the angle efg of the triangle e fg is greater than its angle eg f, and that the greater (i. 19) side is opposite to the greater angle; the side eg is therefore greater than the side ef; but eg is equal to bc; and therefore also bc is greater than ef. Therefore, if two triangles, &c. Q. E. D. PROPOSITION XXV.—THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one greater than the base of the other; the angle also contained by the sides of that which has the greater base shall be greater than the angle contained by the sides equal to them of the other. LET a b c d e f be two triangles which have the two sides a b, a c equal to the two sides de df, each to each, viz. a b equal to de, and ac to df; but the base b c greater than the base ef; the angle bac is likewise greater than the angle edf. For, if it be not greater, it must either be equal to it, or less ; but the angle bac is not equal to the angle edf; because then the base b c would ка be equal (i. 4) to ef; d but it is not; therefore the angle bac is not equal to the angle edf; neither is it less : because then the base bo would be less (i. 24) than the base e f; but it is not; therefore the angle bac is not less than the angle edf; b sf and it was shewn that it is not equal to it; therefore the angle bac is greater than the angle ed f. than the angle edf. Wherefore, if two triangles, &c. Q. E. D. PROPOSITION XXVI.—THEOREM. If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz. either the sides adjacent to the equal angles, or the sides opposite to equal angles in each; then shall the other sides be equal, each to each, and also the third angle of the one to the third angle of the other. Let a bc, def be two triangles which have the angles a b c bca equal to the angles def, efd, viz. a b c to def, and bca to efd; also one side equal to one d side; and first let those sides be equal which are adjacent to the angles that are equal in the two triangles : viz. bc to ef; the other sides shall be equal each to each, viz. a b to de, and ac b f to df, and the third angle bac to the third angle e df. For, if a b be not equal to de, one of them must be the greater. Let ab be the greater of the two, and make bg equal to de, and join gc; therefore, because bg is equal to de, and b c to ef, the two sides g bbc are equal to the two de, ef, each to each ; and the angle gbc is equal to the angle def; therefore the base gc is equal (i. 4) to the base d f, and the triangle gb c to the triangle de f, and the other angles to the other angles, each to each, to which the equal sides are opposite ; therefore the angle gcb is equal to the angle dfe; but dfe is, by the hypothesis, equal to the angle bca; wherefore also the angle bcg is equal to the angle bca, the less to the greater, which is impossible ; therefore a b is not unequal to de, that a is, it is equal to it; and b c is equal to ef; therefore the two a b b c are equal to the two de, ef, each to each; and the angle a b c is equal to the angle def; the base therefore a c is equal (i. 4) to the base d f, and the third angle bac to the third angle ed f. Next, let the sides which are opposite to equal angles in each triangle be equal to one another, viz. a b to de; likewise in this case, the other sides shall be equal, ac to df, and bc to ef; and also the third angle bac to the third ed f. a d b h f For, if b c be not equal to ef, let bc be the greater of them, and make bh equal to ef, and join a h; and because bh is equal to ef, and a b to de; the two a b bh are equal to the two de, ef, each to each ; and they contain equal angles; therefore the base a h is equal to the base df, and the triangle a bh to the triangle def, and the other angles shall be equal, each to each, to which the equal sides are opposite ; therefore the angle bha is equal to the angle efd; but efd is equal to the angle bca; therefore also the angle bha is equal to the angle b ca, that is, the exterior angle bha of the triangle a hc is equal to its interior and opposite angle bca; which is impossible (i. 16); wherefore bc is not unequal to ef, that is, it is equal to it; and a b is equal to de; therefore the two, a b, b c are equal to the two de, ef, each to each ; and they contain equal angles ; wherefore the base a c is equal to the base d f, and the third angle bac to the third angle edf. Therefore, if two triangles, &c. Q. E. D. PROPOSITION XXVII.-THEOREM. If a straight line falling upon two other straight lines makes the alternate angles equal to one another, these two straight lines shall be parallel. Let the straight line e f, which falls upon the two straight lines a b, cd, make the alternate angles a ef, efd equal to one another; a b is parallel to cd. For, if it be not parallel, a b and cd being produced shall meet either towards b, d, or towards a, c: let them be produced and meet towards b, d in the point g; therefore gef is a triangle, and its exterior angle a ef is greater (i. 16) than the interior and opposite angle efg; but it is also equal to it, which is impossible ; therefore a b and cd being produced do not meet towards b, d. In like manner it may be demonstrated, that they do not meet towards a, c, but those straight lines which meet neither way, though produced ever so far, are parallel (35 def.) to one another. a b therefore is parallel to cd. Wherefore, if a straight line, &c. Q. E. D. a PROPOSITION XXVIII. THEOREM. If a straight line falling upon two other straight lines makes the exterior angle equal to the interior and opposite upon the same side of the line, or makes the interior angles upon the same side together equal to two right angles ; the two straight lines shall be parallel to one another. LET the straight line e f, which falls upon the two straight lines a b, cd, make the exterior angle e gb equal to the interior and opposite angle ghd upon the same side ; or b make the interior angles on the same side bgh, ghd together equal to two right angles ; ab is parallel to cd. C h d Because the angle eg b is equal to the f angle g hd, and the angle eg b equal (i. 15) to the angle a gh, the angle å gh is equal to the angle ghd; and they are the alternate angles ; therefore a bis parallel (i. 27) to cd. Again, because the angles bgh, ghd are equal (by hyp.) to two right angles ; and that a gh, b gh, are also equal (i. 13) to two right angles:; the angles a g h, bgh, are equal to the angles bgh, ghd: take away the common angle bgh; therefore the remaining angle agh is equal to the remaining angle ghd; and they are alternate angles; therefore a b is parallel to cd. Wherefore, if a straight line, &c. Q. E. D. ز a 60 PROPOSITION XXIX.—THEOREM. If a straight line fall upon two parallel straight lines, it makes the alternate angles equal to one another ; and the exterior angle equal to the interior and opposite upon the same side ; and likewise the two interior angles upon the same side together equal to two right angles. LET the straight line e f fall upon the parallel straight lines a b c d ; the a alternate angles a gh, ghd, are equal to one another ; and the exterior angle eg b is equal to the interior and opposite, upon the same side ghd; and the two interior angles bgh, g hd upon the same side, are together equal to two right angles. For, if a gh be not equal to ghd, one of them must be greater than the other ; -b let a g h be the greater ; and because the angle a g h is greater than the angle ghd, add to each of them the angle b gh; therefore the angles a gh, bg h are greater than h d the angles bg h, ghd; but the angles agh, bgh, are equal (i. 13) to two right f angles ; therefore the angles bgh, ghd, are less than two right angles ; but those straight lines which, with another straight line falling upon them, make the interior angles on the same side less than two right angles, do meet (12 ax:) together if continually produced ; therefore the straight lines ab, cd, if produced far enough, shall meet ; but they never meet, since they are parallel by the hypothesis ; therefore the angle agh is not unequal to the angle g h d, that is, it is equal to it ; but the angle a gh is equal (i. 15) to the angle eg b; therefore likewise eg b is equal to ghd; add to each of these the angle bgh; therefore the angles eg b, b g h are equal to the angles bgh, ghd; but eg b, bg h are equal (i. 13) to two right angles; therefore also bgh, ghd are equal to two right angles. Wherefore, if a straight line, &c. Q. E. D. с PROPOSITION XXX.—THEOREM.. Straight lines which are parallel to the same straight line are parallel to each other. LET a b, cd be each of them parallel to ef; ab is also parallel to cd. Let the straight line ghk cut a b, e f, cd; and because ghk cuts the parallel straight lines a b, ef, the angle agh is equal (i. 29) to the angle g hf. Again, be g a b cause the straight line gk cuts the parallel straight lines e f, cd, the angle ghfis equal h f (i. 29) to the angle gkd ; and it was shewn that the angle agk is equal to the angle k d ghf; therefore also a gk is equal to gk d ; and they are alternate angles ; therefore a b is parallel (i. 27) to cd; Wherefore, straight lines, &c. Q. E. D. e |