tude; and the two triangles ADE, CDE, on the triangle ADE: bde: ad: de, B But BDE is CDE; and equals must have to equals the same ratio; therefore AD: DB:: AE: EC. Q. E. D. Corol. Hence, also, the whole lines AB, AC, are proportional to their corresponding proportional segments (corol. th. 66). A Line which Bisects any Angle of a Triangle, divides the opposite Side into Two Segments, which are Proportional to the two other Adjacent Sides. Let the angle ACB, of the triangle ABC, be bisected by the line CD, making the angler equal to the angle s: then will the segment AD be to the segment DB, as the side AC is to the side CB. AD DB AC: CB. Or, For, let BE be parallel to co, meeting AC produced at E. E A D B Then, because the line вC cuts the two parallels CD, BE, it makes the angle CBE equal to the alternate angle s (th. 12), and therefore also equal to the angle r, which is equal to s by the supposition. Again, because the line AE cuts the two parallels DC, BE, it makes the angle E equal to the angler on the same side of it (th. 14). Hence, in the triangle BCE, the angles в and E, being each equal to the angle r, are equal to each other, and consequently their opposite sides CB, CE, are also equal (th. 3). But now, in the triangle ABE, the line co, being drawn parallel to the side BE, cuts the two other sides AB, AE proportionally (th. 82), making AD to DB, as is Ac to c or to its equal CB. Q. E. D. THEOREM THEOREM LXXXIV. Equiangular Triangles are similar, or have their Like Sides Proportional. LET ABC, DEF, be two equiangular triangles, having the angle A equal to the angle D, the angle в to the angle E, and consequently the angle c to the angle ; then will AB AC DE : DF. A. B For, make DGAB, and DHAC, and join GH. Then the two triangles ABC, DGH, having the two sides AB, AC, equal to the two DG, DM, and the contained angles A and D also equal, are identical, or equal in all respects (th. 1), namely the angles B and c are equal to the angles & and H. But the angles B and c are equal to the angles E and by the hypothesis; therefore also the angles G and H are equal to the angles E and F (ax. 1), and consequently the line GH is parallel to the side EF (cor. 1, th. 14). Hence then, in the triangle DEF, the line GH, being parallel to the side EF, divides the two other sides proportionally, making DG DH: DE: DF (cor. th. 82.) But DG and DA are equal to AB and AC; therefore also AB AC THEOREM LXXXV. : DE DF. Q. E. D. Triangles which have their Sides Proportional, are Equiangular. In the two triangles ABC, DEF, if AB :: DE: AC: DF: BC EF; the two triangles will have their corresponding angles equal. For, if the triangle ABC be not equiangular with the triangle DEF, suppose some other triangle, as DEG, to be equiangular with ABC. But this is impossible: for if the two triangles ABC, DEG, were equiangular, their sides would be proportional GF (th. 84). So that, AB being to DE as AC to DG, and AB to DE as BC to EG. it follows that DG and EG being fourth proportionals to the same three quantities as well as the two DF, EF, the former VOL. I. 43 former DG, EG, would be equal to the latter, DF, ef. Thus then, the two triangles, DEF DE, having their three sides equal, would be identical (th. 5); which is absurd, since their angles are unequal. THEOREM LXXXVI. Triangles, which have an Angle in the one Equal to an Angle in the other, and the Sides about these angles Proportional, are Equiangular. LET ABC, DEF, be two triangles, having the angle = the angle D, and the sides AB, AC, proportional to the sides DE. DF: then will the triangle ABC be equiangular with the triangle DEf. For, make DG=AB, and Dн=AC, and join GH. A D B F H GE Then, the two triangles ABC, DGH, having two sides equal, and the contained angles A and D equal, are identical and equiangular (th. 1), having the angles G and H equal to the angles B and C. But, since the sides DG, DH, are proportional to the sides DE, DF, the line GH is parallel to Er (th. 82); hence the angles E and F are equal to the angles G and H (th. 14), and consequently to their equals в and c. THEOREM LXXXVII. Q. E. D. In a Right-Angled Triangle, a Perpendicular from the Right Angle, is a Mean Proportional between the Segments of the Hypothenuse; and each of the Sides, about the Right Angle, is a Mean Proportional between the Hypothenuse and the adjacent segment. Let ABC be a right-angled triangle, and CD a perpendicular from the right angle c to the hypothenuse AB; then will A CD be a mean proportional between AD and DB; Bс a mean proportional between AB and BD. D B Or, AD CD CD: DB; and AB: BC BC BD; and AB : AC AC. For For, the two triangles ABC, ADc, having the right angles at c and D equal, and the angle a common, have their third angles equal, and are equiangular (corol. 1, th. 17). In like manner, the two triangles ABC, BDC, having the right angles at c and D equal, and the angle в common, have the third angles equal, and are equiangular. Hence then, all the three triangles ABC, ADC, BDC, being equiangular will have their like sides proportional (th. 84). viz. AD CD CD: DB; Q. E. D. Corol. Because the angle in a semicircle is a right angle (th. 52); it follows, that if, from any point c in the periphery of the semicircle, a perpendicular be drawn to the diameter AB; and the two chords CA, CB, be drawn to the extremities of the diameter: then are AC, BC, CD, the mean proportionals as DB; AC2 = in this theorem, or (by th. 77), CD3 AB. AD; and BC2 AB. BD. =AD. THEOREM LXXXVIII. Equiangular or Similar Triangles, are to each other as the Squares of their Like Sides. Let ABC, DEF, be two equiangular triangles, AB and DE being two like sides; then will the triangle ABC be to the trias the angle DEF, square of AB is to the square of DE, or as AB2 to DE2. C B D H K L M N For, let AL and DN be the squares on AB and DE; also draw their diagonals BK, EM, and the perpendiculars CG, FH, of the two triangles. Then, since equiangular triangles bave their like sides proportional (th. 84), in the two equiangular triangles ABC, DEF, the side AC: DF: AB: DE; and in the two ACG, DFH, the side AC DF: CG: FH; therefore, by equality co: FH ::AB: DE, or CG: AB FH DE. But because triangles on equal bases are to each other as their altitudes, the triangles ABC, ABK, on the same base AB, are to each other, as their altitudes CG, AK, or AB : and and the triangles DEF, DEM, on the same base DE, are as their altitudes F. DM, or DE ; that is, triangle ABC; triangle ABK :: CG: AB, and triangle DEF : triangle DEи ¦¦ But it has been shown that CG: AB theref. of equality AABC: AABK :: or alternately, as ▲ ABC: A DEF:: FH: DE. FH: DE ; But the squares AL, DN, being the double of the triangles ABH, DEM, have the same ratio with them; therefore the AABC ADEF square AL square DN. Q. E. D. THEOREM LXXXIX. All Similar Figures are to each other, as the Squares of their For, draw BE, BD, GK, GI, dividing the figures into an equal number of triangles, by lines from two equal angles B and G. The two figures being similar (by suppos.), they are equiangular, and have their like sides proportional (def. 70). Then, since the angle A is the angle F, and the sides AB, AE, proportional to the sides FG, FK, the triangles ABE, FGK, are equiangular (th. 86). In like manner, the two triangles BCD, GHI, having the angle c = the angle =, and the sides BC, CD, proportional to the sides GA, HI, are also equiangular. Also, if from the equal angles AED, FRI, there be taken the equal angles AEB, FKG, there will remain the equals BED, GKI; and if from the equal angles CDE, HIK, be taken away the equals CDB, HIG, there will remain the equals BDE, GIK; so that the two triangles BDE, GIK, having two angles equal, are also equiangular. Hence each triangle of the one figure, is equiangular with each corresponding triangle of the other. But equiangular triangles are similar, and are proportional to the squares of their like sides (th. 88). Therefore |