Let ABCDE, FGHIK, be two similar polygons. From any angle A, in the polygon ABCDE, 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. H 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, ABC=FGH, BCA=GHF, ACD=FHI: hence BCD= GHI, likewise CDE=HIK, &c. Moreover we shall have AB: FG:: BC: GH :: AC: FH:: CD: HI, &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 as the homologous sides: and the areas are to each other as the squares described on those sides. &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 II. Prop. X.). Secondly. Since the triangles ABC, FGH are similar, we shall have the triangle ABC: FGH AC2: 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 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 homologous 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. PROPOSITION XXVIII. THEOREM. 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 C-B; 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 secants be drawn terminating in the concave arc, the whole secants will be reciprocally proportional to their external segments. 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. XVIII. Cor. 1.); these triangles are therefore similar; and their homologous sides give the proportion, OB OC :: OD: OA. Cor. Hence the rectangle OA.OB is equal to the rectangle OC.OD. Scholium. This proposition, it may be observed, bears a 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. PROPOSITION XXX. THEOREM. If from the same point without a circle, a tangent and a secant be drawn, the tangent will be a mean proportional between the secant and its external segment. From the point O, let the tangent OA, and the secant OC be be drawn then will OC OA: OA: OD, or OA2=OC.OD. For, drawing AD and AC, the triangles O OAD, OAC, have the angle O common; also the angle OAD, formed by a tangent and at chord, has for its measure half of the arc AD (Book III. Prop. XXI.); and the angle Chas the same measure: hence the angle OAD= A C; therefore the two triangles are similar, and we have the proportion OC: OA :: AO OD, which gives OA2=OC.OD. PROPOSITION XXXI. THEOREM. If either angle of a triangle be bisected by a line terminating in the opposite side, the rectangle of the sides including the bisected angle, is equivalent to the square of the bisecting line together with the rectangle contained by the segments of the third side. In the triangle BAC, let AD bisect the angle A; then will AB.AC AD2+BD.DC. Describe a circle through the three points A, B, C ; produce AD till it meets the circumference, and draw CE. = D E C The triangle BAD is similar to the triangle EAC; for, by hypothesis, the angle BAD EAC; also the angle B=E, since they are both measured by half of the arc AC; hence these triangles are similar, and the homologous sides give the proportion BA: AE :: AD: AC; hence BA.AC-AE.AD; but AE=AD+DE, and multiplying each of these equals by AD, we have AE.AD=AD2+ AD.DE; now AD.DE=BD.DC (Prop. XXVIII.); hence, finally, BA.AC AD2+BD.DC. PROPOSITION XXXII. THEOREM In every triangle, the rectangle contained by two sides is equivalent to the rectangle contained by the diameter of the circumscribed circle, and the perpendicular let fall upon the third side. In the triangle ABC, let AD be drawn perpendicular to BC; and let EC be the diameter of the circumscribed circle; then will AB.AC AD.CE. For, drawing AE, the triangles ABD, AEC, are right angled, the one at D, the E other at A: also the angle B=E; these triangles are therefore similar, and they give the proportion AB : CE :: AD: AC; and hence AB.AC-CE.AD. Cor. If these equal quantities be multiplied by the same quantity BC, there will result AB.AC.BC=CE.AD.BC; now AD.BC is double of the area of the triangle (Prop. VI.); therefore the product of three sides of a triangle is equal to its area multiplied by twice the diameter of the circumscribed circle. The product of three lines is sometimes called a solid, for a reason that shall be seen afterwards. Its value is easily conceived, by imagining that the lines are reduced into numbers, and multiplying these numbers together. Scholium. It may also be demonstrated, that the area of a triangle is equal to its perimeter multiplied by half the radius of the inscribed circle. For, the triangles AOB, BOC, AOC, which have a common vertex at O, have for their common altitude the radius of the inscribed circle; hence the sum of these triangles will be equal to the sum of the bases AB, BC, AC, multiplied by half the radius D B F E OD; hence the area of the triangle ABC is equal to the perimeter multiplied by half the radius of the inscribed circle. |