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angle cad (i. 29): but cad is equal to the angle dae (hyp.); therefore also dae is equal to the angle acf. Again, because the straight line fa e

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meets the parallels a d, fc, the outward angle dae is equal to the inward and opposite angle cfa: but the angle a cf has been proved equal to the angle dae; therefore also the angle acf is equal to the angle cfa, and consequently the side af is equal to the side dac (i. 6) and because ad is parallel to fc, a side of the triangle bcf, bd is

to dc, as ba to af (vi. 2); but af is equal to a c; as therefore bd is to dc, so is ba to a c.

Let now bd be to dc as ba to ac, and join ad; the angle cad is equal to the angle da e

The same construction being made, because bd is to dc as ba to a c; and that bd is also to dc, as ba to af (vi. 2); therefore ba is to ac as ba to af (v. 11); wherefore a c is equal to af (v. 9), and the angle afc equal (i. 5) to the angle a cf: but the angle afc is equal to the outward angle e a d, and the angle a cf to the alternate angle cad; therefore also ead is equal to the angle cad. Wherefore, if the outward, &c. Q. E. D.

PROPOSITION IV.-THEOREM.

The sides about the equal angles of equiangular triangles are proportionals; and those which are opposite to the equal angles are homologous sides, that is, are the antecedents or consequents of the ratios.

LET abc, dce be equiangular triangles, having the angle abc equal to the angle dce, and the angle acb to the angle dec, and consequently (i. 32) the angle bac equal to the angle cde. The sides about the equal angles of the triangles abc, dce are proportionals; and those are the homologous sides which are opposite to the equal angles.

Let the triangle dce be placed so that its side ce may be contiguous to bc, and in the same straight line with it: and because the angles a bc,

acb, are together less than two right angles (i. 17), a b c and d e c, which is equal to a cb, are also less than two right angles; wherefore ba, ed produced shall meet (i. ax. 12); let them be produced and meet in the point f; and because the angle abc is equal to the angle dce, bf is parallel to c d (i. 28). Again, because the angle a cb is equal to the angle dec, ac is parallel to fe (i. 28): therefore fa cd is a parallelogram, and consequently af is equal to cd, and ac to fd (i. 34): and because a c is parallel to fe, one of the sides of the triangle fb e, ba is to afas bc to ce (vi. 2): but af is equal to cd; therefore (v. 7), as ba to cd, so is bc to ce; and alternately, as ab to bc, so is dc to ce

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Again, because cd is parallel to bf, as bc to ce, so is fd to de (vi. 2); but fd is equal to a c; therefore, as bc to ce, so is a c to de: and alternately, as bc to ca, so is ce to ed: therefore, because it has been proved that a b is to bc, as dc to ce, and as bc to ca, so is ce to ed, ex æquali (v. 22), ba is to ac, as cd to de. Therefore the sides, &c. Q. E. D.

PROPOSITION V.-THEOREM.

If the sides of two triangles about each of their angles be proportionals, the triangles shall be equiangular, and have their equal angles opposite to the homologous sides.

LET the triangles abc, def have their sides proportionals, so that ab is to be as de to ef; and be to ca as efto fd; and consequently, ex aequali, ba to ac as ed to df; the triangle abc is equiangular to the triangle def, and their equal angles are opposite to the homologous sides, viz. the angle abc equal to the angle def, and bca to efd, and also bac to edf.

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At the points e, f, in the straight line ef, make (i. 23) the angle feg equal to the angle a bc, and the angle efg equal to bca; wherefore the remaining angle bac is equal to the remaining angle egf (i. 32), and the triangle abc is therefore equiangular to the triangle gef; and consequently they have their sides opposite to the equal angles proportionals (vi. 4). Wherefore as a b to bc, so is ge to ef; but as a b to bc, so is de to ef; therefore as de to ef, so is ge to ef (v. 11) therefore de and have the same ratio to ef, and consequently are equal (v. 9): for the same reason, d f is equal to fg: and because, in the triangles def gef de is equal to eg, and ef common' the two sides de, ef, are equal to the two ge, ef, and the base dfis equal to the base gf; therefore the angle def is equal (i. 8) to the angle gef, and the other angles to the other angles which are subtended by the equal sides (i. 4). Wherefore the angle dfe is equal to the angle gfe, and edf to egf and because the angle def is equal to the angle gef, and gef to the angle a bc; therefore the angle a bc is equal to the angle def. For the same reason, the angle a cb is equal to the angle dfe, and the angle at a to the angle at d. Therefore the triangle a bc is equiangular to the triangle def. Wherefore, if the sides, &c. Q. E. D.

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PROPOSITION VI.-THEOREM.

If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, the triangles shall be

equiangular, aad shall have those angles equal which are opposite to the homologous sides.

LET the triangles ab c, def have the angle bac in the one equal to the angle e d f in the other, and the sides about those angles proportionals; that is, ba to a c, as ed to df; the triangles a bc, def are equiangular, and have the angle a b c equal to the angle def, and a cb to dfe.

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At the points d, f, in the straight line df, make (i. 23) the angle fdg equal to either of the angles bac, edf; and the angle d fg equal to the angle acb wherefore the remaining angle at b is equal to the remaining one at g (i. 32), and consequently the triangle abc is equiangular to the triangle dgf; and therefore as ba to a c, so is (vi. 4) gd to df; but by the hypothesis, as ba to a c, so is ed to df; as therefore ed to df, so is (v. 11) gd to df; wherefore ed is equal (v. 9) to dg; and df is common to the two triangles edf, gdf: therefore the two sides ed, df are equal to the two sides gd, df; and the angle e df is equal to the angle gdf; wherefore the base ef is equal to the base fg (i. 4) and the triangle edf to the triangle gdf, and the remaining angles to the remaining angles, each to each, which are subtended by the equal sides; therefore the angle dfg is equal to the angle dfe, and the angle at g to the angle at e: but the angle dfg is equal to the angle a cb, therefore the angle a cb is equal to the angle dfe and the angle ba c is equal to the angle e d f (hyp.); wherefore also the remaining angle at b is equal to the remaining angle at e. Therefore the triangle abc is equiangular to the triangle def. Wherefore, if two triangles, &c. Q. E. D.

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PROPOSITION VII.-THEOREM.

If two triangles have one angle of the one equal to one angle of the other, and the sides about two other angles proportionals, then if each of the remaining angles be either less, or not less, than a right angle, or if one of them be a right angle, the triangles shall be equiangular, and have those angles equal about which the sides are proportionals.

LET the two triangles abc, def have one angle in the one equal to one angle in the other, viz. the angle bac to the angle e df, and the sides about two other angles abc, def proportionals, so that a b is to be as de to ef; and in the first case, let each of the remaining angles at c, f be less than a right angle. The triangle abc is equiangular to the triangle def, viz. the angle abc is equal to the angle def, and the remaining angle at c to the remaining angle at f

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For if the angles a b c, d e f be not equal, one of them is greater than the other let abc be the greater, and at the point b, in the straight line ab, make the angle abg equal to the angle (i. 23) def and because the angle at a is equal to the angle at d, and the angle a bg to the angle def; the remaining angle a gb is equal (i. 32) to the remaining

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angle dfe: therefore the triangle abg is equiangular to the triangle def; wherefore (vi. 4) as ab is to bg, so is de to ef, but as de to ef, so, by hypothesis, is ab to bc; therefore as ab to be, so is ab to bg (v. 11): and because ab has the same ratio to each of the lines bc, bg; bc is equal (v. 9) to bg, and therefore the angle bg c is equal to the angle bcg (i. 5): but the angle bcg is, by hypothesis, less than a right angle; therefore also the angle bgc is less than a right angle, and the adjacent angle a gb must be greater than a right angle (i. 13). But it was proved that the angle a gb is equal to the angle at f; therefore the angle at f is greater than a right angle: but, by the hypothesis, it is less than a right angle; which is absurd. Therefore the angles abc, def are not unequal, that is, they are equal and the angle at a is equal to the angle at d; wherefore the remaining angle at c is equal to the remaining angle at f: therefore the triangle abc is equiangular to the triangle def.

Next, let each of the angles at c, f be not less than a right angle: the triangle a bc is also in this case equiangular to the triangle de f

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The same construction being made, it may be proved in like manner that bc is equal to bg, and the angle at c equal to the angle bgc: but the angle at c is not less than a right angle; b therefore the angle bgc is not less than a right angle: wherefore two angles of the triangle bgc are together not less than two right angles, which is impossible (i. 17); and therefore the triangle abc may be proved to be equiangular to the triangle def, as in the first case.

Lastly, let one of the angles at c, f, viz. the angle at c, be a right angle; in this case likewise the triangle abc is equiangular to the triangle def.

For if they be not equiangular, make, at the point b of the straight line ab, the angle abg equal to the angle def; then it may be proved, as in the first case, that bg is equal to bc but the angle bcg is a right angle, therefore (i. 5) the angle bgc is also a right angle; whence two of the angles of the triangle bgc are together not less than two right angles, which is impossible (i. 17): therefore the triangle abc is equiangular to the triangle def. Wherefore, if two triangles, &c. Q. E. D.

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PROPOSITION VIII.-THEOREM.

In a right-angled triangle, if a perpendicular be drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle and to one another.

LET abc be a right-angled triangle, having the right angle bac; and from the point a let ad be drawn perpendicular to the base bc: the triangles a bd, adc are similar to the whole triangle abc, and to one another.

Because the angle bac is equal to the angle adb, each of them being a right angle, and that the angle at b is common to the two trian

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gles abc, abd; the remaining angle a cb is equal to the remaining angle bad (i. 32) therefore the triangle abc is equiangular to the triangle a bd, and the sides about their equal angles are proportionals (vi. 4); wherefore the triangles are similar (vi. def. 1): in the like manner it may be demonstrated, that the triangle a dc is equiangular and similar to the triangle abc: and the triangles abd, a cd, being both equiangular and similar to abc, are equiangular and similar to each other. Therefore, in a right-angled triangle, &c. Q. E. D.

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COR. From this it is manifest that the perpendicular drawn from the right angle of a right-angled triangle to the base, is a mean proportional between the segments of the base and also, that each of the sides is a mean proportional between the base, and its segment adjacent to that side because in the triangles bda, a dc, bd is to da, as da to dc (vi. 4); and in the triangles abc, dba, bc is to ba, as ba to bd (vi. 4); and in the triangles abc, acd, bc is to ca, as ca to cd (vi. 4).

PROPOSITION IX.-PROBLEM.

From a given straight line to cut off any part required.

LET ab be the given straight line; it is required to cut off any part from it.

From the point a draw a straight line a c, making any angle with a b; and in ac take any point d, and take ac the same multiple of a d, that ab is of the part which is to be cut off from it: join bc, and draw de parallel to it: then ae is the part required to be cut off.

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Because ed is parallel to one of the sides of the triangle a bc, viz. to bc, as cd is to da, so is be to ea (vi. 2); and by composition (v. 18), ca is to ad, as ba to ae: but ca is a multiple of a d; therefore (v. D) ba is the same multiple of ae: whatever part therefore ad is of a c, ae is the same part of ab: wherefore, from the straight line ab the part required C is cut off. Which was to be done.

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