LAEG is the LACB; .. also (Ax. 1) the LABC is = the DEF, and the LACB is the LDFE. Hence the As are equiangular, and ... similar. (Prop. 61, cor.) PROPOSITION LXIV.-THEOREM. Equal parallelograms, AB, BC, which have one angle FBD of the one, equal to one angle EBG of the other, have the sides about the D equal angles reciprocally proportional; and parallelograms which have one angle of the one, equal to one angle of the other, and the B sides about the equal angles reciprocally proportional, (viz. DB: BE=GB: BF), are equal. Let the sides DB, BE, be placed in the same straight line; then since the LDBF the LEBG, add the FBE to each; .. the two Ls DBF+FBE, are the Ls GBE+ FBE; but DBF+FBE, are = two Ls, since DBE is a straight line, (Prop. 1); .. the Ls GBE+FBE, are together equal to two Ls, and hence (Prop. 2) GBF is a straight line. Now since AB=0BC, and FE is another DAB:OFE=OBC: FE. (Lem. p. 116.) But OAB: □FE-DB: BE, and BČ: FE =GB: BF, (Prop. 58), and .. DB: BE=GB: BF. Next, let the sides about the = Ls be reciprocally proportional, that is, let DB: BE=GB: BF, the AB will the FE;. For DB: BE=GB: BF, and since by (Prop. 58), DB: BE=AB:OFE, and GB: BF=BC: FE;.. AB: FE=BC: FE. (Alg. 102.) That is, the s AB and BC have the same ratio to AB=BC. (Lem. p. 116.) Cor. 1. Since triangles are the halves of parallelograms, upon the same base, and having the same altitude, equal triangles, which have one angle of the one equal to one angle of the other, have the sides about these angles reciprocally proportional; and if two triangles have one angle of the one, equal to one angle of the other, and the sides about these angles reciprocally proportional, the triangles are equal. Cor. 2. If the angle DBF were a right angle, the parallelograms would be rectangles; hence the sides of twa I equal rectangles can be converted into a proportion, by making the sides of the one the extremes, and the sides of the other the means; and if four straight lines be proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means. Cor. 3. If the means of the proportion be equal, the rectangle formed by them will be a square. Hence, "when three straight lines are proportional, the rectangle contained by the extremes is equal to the square on the mean;" and conversely. PROPOSITION LXV.-THEOREM. If two chords, AB, CD, cut one another in a point E, within a circle ACD, the rectangle contained by the segments of the one shall be equal to the rectangle contained by the segments of the other. That is, the rectangle AE·EB-CE.ED. Join CA and BD, then the LCAE is the LBDE; for they stand on the B same arc CB, and the CEA is = the [BED, (Prop. 3); hence the As AEC, BED, are equiangular, (Prop. 19, cor. 1); hence AE: EC-DE: EB, (Prop. 61.) And.. AE.EB=CE-ED. (Prop. 64, cor. PROPOSITION LXVI. THEOREM. If from a point E, without a circle ABC, two straight lines be drawn, cutting the circle, the rectangle contained by the whole AE, and the external segment, EB of the one, will be equal to the rectangle contained by the whole, EC, and the external segment, ED, of the A other. That is, AE EB-CE.ED. Join AC and BD, then the LEDB is 2.) B = the LEAC, (Prop. 48, cor.), and the LE is common; .. the As EBD, ECA, are equiangular, (Prop. 19, cor. 1); hence AE:EC=DE: EB, (Prop. 61.) And.. AE·EB⇒CE-ED. (Prop. 64, cor. 2.) PROPOSITION LXVII. THEOREM. If from a point A without a circle, there be drawn a secant AB, and a tangent AD, the rectangle contained by the whole secant AB, and its external segment AC, will be equal to the square on the tangent AD. That is, AB⋅AC=AD2. Join DB and DC, then the LADC (contained by the tangent AD and the chord DC) is the LDBA, (Prop. 55), and the LA is common to the two As ADC, ABD. D B ..these As are equiangular, (Prop. 19, cor. 1); and hence BA: AD=AD: AC. (Prop. 61.). .. AD-AC-AB. (Prop. 64, cor. 3.) PROPOSITION LXVIII. THEOREM. If a point E be taken in the circumference of a circle, and straight lines AE, EB be drawn to the extremities of the P/ diameter AB, and also EC perpendicular to the diameter, then AE is a mean próportional between the diameter and the adjacent segment AC; EB is a mean proportional between the diameter and the adjacent segment BC; and EC is a mean proportional between the segments of the diameter AC, CB. Or AE2 =ABAC, EC-AC CB, and BE2-AB-BC. B For the As AEB, ACE, are equiangular, since the Ls AEB, ACE, are Ls, and the ZA common, (Prop. 19, cor. 1); also the As AEB, ECB, are equiangular, since they have the Ls AEB, ECB, Ls, and the LB common, (Prop. 19, cor. 1); hence also the As ACE, ECB, are equiangular, since each of them has been shown to be equiangular to AAEB. Now, since the As BAE, EAC, are equiangular, BA: AE AE: AC; .. AE2=AB.AC. And since the As BCE, ECA, are equiangular, And since the As ABE, EBC, are equiangular, Cor. 1. Since AEB is a right-angled triangle, and EC is drawn from the right angle perpendicular to the hypotenuse; "the triangles on each side of a perpendicular on the hypotenuse, drawn from the right angle of a right angled triangle, are similar to the whole, and to one ano ther; each side is a mean proportional between the hypotenuse and its adjacent segment, and the perpendicular is a mean proportional between the segments of the hypo tenuse." Cor. 2. This furnishes another proof of Proposition 39; for since AE2-AB AC, and BE2= AB BC; ..AE2+BE2=AB AC+AB BC-AB2. (Prop. 35.) PROPOSITION LXIX.-THEorem. Triangles ABC, DBE, which have one angle, ABC, of the one, equal to one angle, DBE, of the other, are to one another in the ratio compounded of the ratios of the sides about the equal angles. That is, ABC: ADBE AB BC: DB.BE. B Let the As ABC, DBE be so placed that AB and BE may be in one straight line, then DB and BC will also be in one straight line; and join AD, then (Prop. 58). AABC: AABD=CB: BD and AABD: ADBE AB: BE; .. compounding these ratios, and omitting from the first and second terms ABD, we have AABC: ADBE AB·BC: DB.BE, (Alg. 116). Cor. Parallelograms which have one angle of the one equal to one angle of the other, are to one another in the ratio which is compounded of the ratios of the sides about the equal angles. For they are double of the As of which the property has just been demonstrated. PROPOSITION LXX.-THEOREM. Similar triangles ABC, DEF, are to one another as the squares of their corresponding sides. For by last Theorem we have, B the As are equiangular, the LA being AABC: ADEF=AB.AC: DE-DF, but AB: AC=DE: DF, (Prop. 61). .. AB: DE=AC: DF, (Alg. 105), and AB: DE=AB: DE, from equality. .. AB2 : DE2=AB·AC: DE DF, (Alg. 116). : = the LD, ABC:ADEF=AB*:DE”, (Alg. 102). Cor. Equiangular parallelograms are to one another as the squares of their corresponding sides, since they are the doubles of equiangular triangles. other as the squares of their corresponding sides. K Let E and K be corresponding angles, AB and FG corresponding sides, draw the diagonals EB, EC in the one, and KG, KH in the other; and let A and F, B and G, C and H, and D and I be the other corresponding angles. Since the polygons are similar, and A and F equal Ls, .. AB: AE FG: FK, and hence (Prop. 63), the As ABE, FGK are equiangular; .. the LABE the FGK; take away these equal Ls from the equal Ls ABC, FGH, and there remain the equal Ls CBE, HGK. Now since the AABE is similar to the AFGH,.. AB: BE=FG: GK, and the polygons being similar, AB : BC=FG: GH; ·. (Alg. 111), EB: BC=KG: GH; but the Ls CBE, HGK were proved equal, .. the ABCE is similar to the GHK; hence the Ls BCE, GHK are equal: but the Ls BCD, GHI are also equal; .. the remaining Ls ECD, KHI are equal. Again, since the As BCE, GHK are similar, .. BC: CE=GH: HK; and since the polygons are similar, BC: CD=GH: HI; .. (Alg. 111) EC: CD=KH: HI; but the Ls ECD, KHI are equal, .. the ACDE is similar to the AHIK, (Prop. 63.) Since the As ABE, FGK are similar, we have AABE: AFGK=BE2 : GK2, (Prop. 70); and since the As BCE, GHK are similar, we have ABCE: AGHK=BE2: GK2, (Prop. 70). ..▲ABE: FGK=^BCE: ▲GHK, (Alg. 102). By the same mode of reasoning we should find AGHK ACDE: AHIK. BCE: And from this series of equal ratios, (Alg. 113), we conclude that the sum of the antecedents, that is the As ABE, BCE, CDE, or the whole polygon ABCDE is to the sum of all the consequents, that is the As FGK, GHK, HIK, or the whole polygon FGHIK as one ante |