AM equiangular, aad shall have those angles equal which are opposite to the homologous sides. LET the triangles a b c d e f have the angle bac in the one equal to the angle e df in the other, and the sides about those angles proportionals; that is, ba to a c, as ed to df; the triangles a b c d e f are equiangular, and have the angle a b c equal to the angle d e f, and a cb to dfe. 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 dfg equal to the angle acb: wherefore the remaining a angle at b is equal to the remaining one d 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; b с е f wherefore ed is equal (v. 9) to dg; and d f is common to the two triangles edf, gdf: therefore the two sides e d, df are equal to the two sides gd, df; and the angle edf 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 ac b, therefore the angle a cb is equal to the angle dfe: and the angle bac 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 a b c is equiangular to the triangle def. Wherefore, if two triangles, &c. Q. E. D. 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 a b c, 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 a b c d e f 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 a b c 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. For if the angles a b c d e f be not equal, one of them is greater than the other : let a b c be the greater, and at the point b, in the straight line a b, make the angle a bg equal to the angle (i. 23) def: and because the angle at a is equal to the angle at d, and the angle a b.g. the angle def; the remaining angle a gb is equal (i. 32) to the remaining to a ong angle dfe: therefore the triangle abg is equiangular to the triangle def; wherefore (vi. 4) as a b is to bg, so is de to ef, but as de to a ef, so, by hypothesis, is ab to bc; therefore as a b to bc, so is ab to bg (v. 11): and because a b has the same ratio to each of the lines b c bg; bc is equal (v. 9) to bg, and therefore the angle bg cis equal to the angle b f bcg (i. 5); but the angle bog 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 ag b 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 a bc, 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 a bc 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 equi a angular to the triangle d e f. The same construction being made, d it may be proved in like manner that bc is equal to bg, and the angle at c g с f 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 a b c 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. a For if they be not equiangular, make, at the point b of the straight line a b, the angle a bg, equal to the angle def; then it may be proved, as in the first case, that bg is equal to ъ d bc: but the angle bc ğ 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 f together not less than two right angles, which is impossible (i. 17): therefore b the triangle a b c is equiangular to the triangle de f. Wherefore, if two triangles, &c. Q. E. D. с e g a 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 tri angle and to one another. LET a bc be a right-angled triangle, having the right angle bac; and from the point a let ad be drawn perpendicular to the base b c : the triangles a bd, adc are similar to the whole triangle a b c, 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 gles abc, abd; the remaining angle a cb is equal to the remaining angle bad (i. 32): therefore the triangle a b c is equiangular to the triangle a bd, and the sides about their equal angles are pro portionals (vi. 4); wherefore the triangles b d are similar (vi. def. 1): in the like man may be demonstrated, that the triangle a dc is equiangular and similar to the triangle abci and the triangles a b d, a cd, being both equiangular and similar to a b c, are equiangular and similar to each other. Therefore, in a right-angled triangle, &c. Q. E. D. 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 b da, a dc, bd is to da, as d a to do (vi. 4); and in the triangles a b c, d ba, b c is to ba, as ba to bd (vi. 4); and in the triangles a b c, acd, bc is to ca, as ca to cd (vi. 4). с ner it a PROPOSITION IX.-PROBLEM. From a given straight line to cut off any part required. LET a b 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 a c take any point d, and take a c the same multiple of a d, that a b is of the part which is to be cut off from it : join bc, and draw de parallel to it: then a e is the part required to be cut off. d Because ed is parallel to one of the sides of the triangle a b c, 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 a e: but ca is a multiple of a d; therefore (v. D) b a is the same multiple of a e: whatever part therefore a d is of ac, a e is the same part of a b: wherefore, from the straight line a b the part required b C is cut off. Which was to be done. PROPOSITION X.-PROBLEM. a To divide a given straight line similarly to a given divided straight line, that is, into parts that shall have the same ratios to one another which the parts of the divided given straight line have. LET a b be the straight line given to be divided, and a c the divided line : it is required to divide a b similarly to a C Let a c be divided in the points d, e; and let a b, ac be placed so as to contain any angle, and join bc, and through the points d, e, draw (i. 31) df, eg parallels to it; and through d draw dhk parallel to ab: therefore each of the figures fh, h b, is a parallelogram: wherefore dh is equal (i. 34) to fg, and hk to gb: and because he is parallel to k c, one of the sides of the triangle d kc, as ce to ed, so is (vi. 2) kh to hd: but k h is equal to bg, and hd to f d gf; therefore, as ce to ed, so is bg to gf. 8 Again, because fd is parallel to eg, one of the sides of the triangle age, as ed to da, so is gf to fa: but it has been proved that ce is to ed b k as bg to gf; and as ed to da, so is gf to fa : therefore the given straight line ab is divided similarly to a c. Which was to be done. e PROPOSITION XI.- PROBLEM. To find a third proportional to two given straight lines. LET a b, ac be the two given straight lines, and let them be placed so as to contain any angle; it is required to find a third proportional to a b, a c. Produce a b, a c, to the points d, e; and make bd equal to ac; and having joined bc, through d draw de parallel to it (i. 31). b Because bc is parallel to de, a side of the triangle ade, a b is (vi. 2) to bd, as ac to ce: but bd is equal to ac; as therefore ab to ac, so is ac to ce. Wherefore, to the two given straight lines ab, a c, a third proportional ce is found. Which was to be done. d e C a PROPOSITION XII. PROBLEM. To find a fourth proportional to three given straight lines. LET a, b, c be the three given straight lines; it is required to find a fourth proportional to a, b, c. Take two straight lines de, df, containing any angle edf; and upon these make dg equal to a, ge equal to d b, and d h equal to c; and having joined gh, draw ef parallel (i. 31) to it through b the point e: and because g h is parallel to e f, one of the sides of the triangle def, dg is to ge as dh to hf (vi. 2); but dg is equal to a, ge to b, and d'h to c; therefore, as a is to b, so is c to hf. Wherefore to the three given straight lines a, b, c, a fourth propore f tional hf is found. Which was to be done. PROPOSITION XIII.- PROBLEM. To find a mean proportional between two given straight lines. LET a b, bc be the two given straight lines : it is required to find a mean proportional between them. Place a b, b c in a straight line, and upon ac describe the semicircle adc, and from the point b draw (i. 11) d bd at right angles to a c, and join ad dc. Because the angle adc in a simicircle is a right. angle (iii . 31), and because in the right-angled triangle a dc, db is drawn from the right angle perpendicular to the base, db is a mean proportional between a b c a b b c the segments of the base (vi. 3. cor.) : therefore between the two given straight lines a b, bc, a mean proportional d b is found. Which was to done. PROPOSITION XIV.—THEOREM. Equal parallelograms, which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional ; and parallelograms that have one angle of the one equal to one angle of the other, and their sides about the equal angles reciprocally proportional, are equal to one another. LET a b, bc be equal parallelograms, which have the angles at b equal, and let the sides db, be be placed in the same straight line; wherefore are in one straight line (i. 14). The sides of the parallelo also fb, bg |