inequality between the quantities, a train of accurate reasoning would lead us to a manifest and palpable absurdity; from which we are forced to conclude that the two quantitics are equal. Cor. If from a point A, in the circumference of a circle, two chords AB, AC, be drawn to the extremities of a diameter BC, the triangle BAC will be right angled at A (Book III. Prop. BD XVIII. Cor. 2.); hence, first, the perpendicular AD is a mean proportional between the two segments BD, DC, of the diameter, or what is the same, AD2-BD.DC. Hence also, in the second place, the chord AB is a mean proportional between the diameter BC and the adjacent segment BD, or, what is the same, AB- BD.BC. In like manner, we have AC2-CD.BC; hence AB2 : AC2 : : BD: DC: and comparing AB and AC, to BC2, we have AB2 : BC2:: BD : BC, and AC2: BC2: DC: BC. Those proportions between the squares of the sides compared with each other, or with the square of the hypothenuse, have already been given in the third and fourth corollaries of Prop. XI. PROPOSITION XXIV. THEOREM. Two triangles having an angle in each equal, are to each other as the rectangles of the sides which contain the equal angles. In the two triangles ABC, ADE, let the angle A be equal to the angle A; then will the triangle ABC ADE :: AB.AC: AD.AE. Draw BE. The triangles ABE, ADE, having the coinmon vertex E, have the same altitude, and consequently are to each other as their bases (Prop. VI. Cor.): that is, ABE ADE :: AB: AD. In like manner, ABC ABE :: AC : AE. Multiply together the corresponding terms of these proportions, omi the common term ABE; we have ABC ADE : AB.AC: AD.AE. Cor. Hence the two triangles would be equivalent, if the rectangle AB.AC were equal to the rectangle AD.AE, or if we had AB AD: : AE: AC, which would happen if DC were parallel to BE. : PROPOSITION XXV. THEOREM. Two similar triangles are to each other as the on their homologous sides. Let ABC, DEF, be two similar trian- A gles, having the angle A equal to D, and the angle B-E. Then, first, by reason of the equal an- G gles A and D, according to the last proposition, we shall have ABC DEF :: AB.AC: DE.DF. B squares H PROPOSITION XXVI. THEOREM. described D And multiplying the terms of this proportion by the corresponding terms of the identical proportion, AC DF: AC: DF, there will result AB.AC DE.DF:: AC2: DF. Consequently, ABC: DEF : : AC2 : DF2. Therefore, two similar triangles ABC, DEF, are to each other as the squares described on their homologous sides AC, DF, or as the squares of any other two homologous sides. Two similar polygons are composed of the same number of triangles, similar each to each, and similarly situated, Let ABCDE, FGHIK, be two similar polygons. From any angle A, in the polygon ABCDF, B draw diagonals AC, AD to the other angles. From the homologous angle F, in the other polygon FGHIK, draw diagonals FH, FI to the other angles. G F K These polygons being similar, the angles ABC, FGH, which are homologous, must be equal, and the sides AB, BC, must also be proportional to FG, GH, that is, AB FG :: BC: GH (Def. 1.). Wherefore the triangles ABC, FGH, have each an equal angle, contained between proportional sides; hence they are similar (Prop. XX.); therefore the angle BCA is equal to GHF. Take away these equal angles from the equal angles BCD, GHI, and there remains ACD=FHI. But since the triangles ABC, FGH, are similar, we have AC: FH :: BC: GH; and, since the polygons are similar, BC: GH:: CD: HI; hence AC FH : : CD: HI. But the angle ACD, we already know, is equal to FHI; hence the triangles ACD, FHI, have an equal angle in each, included between proportional sides, and are consequently similar (Prop. XX.). In the same manner it might be shown that all the remaining triangles are similar, whatever be the number of sides in the polygons proposed: therefore two similar polygons are composed of the same number of triangles, similar, and similarly situated. Scholium. The converse of the proposition is equally true: If two polygons are composed of the same number of triangles similar and similarly situated, those two polygons will be similar. For, the similarity of the respective triangles will give the angles, ABCFGH, BCA=GHF, ACD=FÍI: hence BCD= GHI, likewise CDE=HIK, &c. Moreover we shall have AB: FG::BC : GH :: ÁC:FH::CD: H1, &c.; hence the two polygons have their angles equal and their sides proportional; consequently they are similar. PROPOSITION XXVII. THEOREM. The contours or perimeters of similar polygons are to each other : First. Since, by the nature of similar figures, B we have AB : FG :: BC GH:: CD: HI, &c. we conclude from this series of equal ratios that the sum of the antecedents AB+BC+CD, A PROPOSITION XXVIII. THEOREM. I E &c., which makes up the perimeter of the first polygon, is to the sum of the consequents FG+GH+HI, &c., which makes up the perimeter of the second polygon, as any one antecedent is to its consequent; and therefore, as the side AB is to its corresponding side FG (Book 11. Prop. X.). Secondly. Since the triangles ABC, FGH are similar, we shall have the triangle ABC: FGH AC: FH2 (Prop. XXV.); and in like manner, from the similar triangles ACD, FHI, we shall have ACD: FHI :: AC2: FH2; therefore, by reason of the common ratio, AC2: FH2, we have ABC FGH :: ACD : FHI. By the same mode of reasoning, we should find ACD FHI ADE: FIK; and so on, if there were more triangles. And from this series of equal ratios, we conclude that the sum of the antecedents ABC+ACD+ADE, or the polygon ABCDE, is to the sum of the consequents FGH+FHI+FIK, or to the polygon FGHIK, as one antecedent ABC, is to its consequent FGH, or as AB2 is to FG2 (Prop. XXV.); hence the areas of similar polygons are to each other as the squares described on the homoiogous sides. Cor. If three similar figures were constructed, on the three sides of a right angled triangle, the figure on the hypothenuse would be equivalent to the sum of the other two: for the three figures are proportional to the squares of their homologous sides; but the square of the hypothenuse is equivalent to the sum of the squares of the two other sides; hence, &c. The segments of two chords, which intersect each other in a circle, are reciprocally proportional. Let the chords AB and CD intersect at 0: then will AO : DO :: OC : OB. Draw AC and BD. In the triangles ACO, BOD, the angles at O are equal, being vertical; the angle A is equal to the angle D, because both are inscribed in the same segment (Book III. Prop. XVIII. Cor. 1.); for the same reason the angle CB; the triangles are therefore similar, and the homologous sides give the proportion AO DO CO: OB. Cor. Therefore AO.OB DO.CO: hence the rectangle under the two segments of the one chord is equal to the rectangle under the two segments of the other. PROPOSITION XXIX. THEOREM. If from the same point without a circle, two secunts be drawn terminating in the concave arc, the whole secants will be reciprocally proportional to their external segments. A Let the secants OB, OC, be drawn from the point 0: then will OB : OC :: OD : OA. For, drawing AC, BD, the triangles OAC, OBD have the angle O common; likewise the angle B C (Book III. Prop. VIII. Cor. 1.); these triangles are therefore similar; and their homologous sides give the proportion, 13 OB : OC :: OD : OA. Cor. Hence the rectangle OA.OB is equal to the rectangle OC.OD. 1 Scholium. This proposition, it may be observed, bears à great analogy to the preceding, and differs from it only as the two chords AB, CD, instead of intersecting each other within. cut each other without the circle. The following proposition: may also be regarded as a particular case of the proposition just demonstrated. |