Similar polygons may be divided into the same number of triangles, similar, each to each, and similarly placed. Let ABCDE and FGHIK be two similar polygons, the angle A being equal to the angle F, B to G, C to H, and so on: then can they be divided into the same number of similar triangles, similarly placed. The Because the polygons are similar, the triangles ABC and FGH have the angles B and G equal, and the sides about these angles proportional; they are, therefore, similar (P. XX.). Since these triangles are similar, we have the angle ACB equal to FHG, and the sides AC and FH, proportional to BC and GH, or to CD and HI. angle BCD being equal to the angle GHI, if we take from the first the angle ACB, and from the second the equal angle FHG, we shall have the angle ACD equal to the angle FHI: hence, the triangles ACD and FHI have an angle in each equal, and the including sides proportional; they are therefore similar. In like manner, it may be shown that ADE and FIK are similar ; which was to be proved. Cor. 1. The corresponding triangles in the two polygons are homologous triangles, and the corresponding diagonals are homologous diagonals. Cor. 2. Any two homologous triangles are like parts of the polygons to which they belong. For, ABC and FGH being similar, we have, ABC: FGH : AC 72 : FH2 ; Cor. 3. If two polygons are made up of similar triangles, similarly placed, the polygons themselves will be similar. The perimeters of similar polygons are to each other as any two homologous sides; and the polygons are to each other as the squares of any two homologous sides. 1o. Let ABCDE and FGHIK be similar polygons: then will their perimeters be to each other as any two homologous sides. polygons are to each other as AB to FG, or as any other two homologous sides; which was to be proved. 2o. The polygons will be to each other as the squares like parts of the polygons to which they belong, the polygons will be to each other as these triangles; but these triangles, being similar, are to each other as the squares of AB and FG hence, the polygons are to each other as the squares of AB and FG, or as the squares of any other two homologous sides; which was to be proved. : Cor. 1. Perimeters of similar polygons are to each other as their homologous diagonals, or as any other homologous lines; and the polygons are to each other as the squares of their homologous diagonals, or as the squares of any of any other homologous lines. Cor. 2. If the three sides of a right-angled triangle be made homologous sides of three similar polygons, these polygons will be to each other as the squares of the sides of the triangle. But the square of the hypothenuse is equal to the sum of the squares of the other sides, and consequently, the polygon on the hypothenuse will be equal to the sum of the polygons on the other sides. PROPOSITION XXVIII. THEOREM. If two chords intersect in a circle, their segments will be reciprocally proportional. Let the chords AB and CD intersect at : then will their segments be reciprocally proportional; that is, one segment of the first will be to one segment of the second, as the remaining segment of the second is to the remaining segment of the first. For, draw CA and BD. Then will the angles ODB and OAC be equal, because each is measured by half of the arc CB (B. III., P. XVIII.). The angles OBD and OCA, will also be equal, because each is measured by B half of the arc AD: hence, the triangles OBD and OCA are similar (P. XIX., C.), and consequently, their homologous sides are proportional: hence, DO : A0 :: OB : OC; which was to be proved. Cor. From the above proportion, we have, DO × OC = A0 × OB; that is, the rectangle of the segments of one chord is equal to the rectangle of the segments of the other. If from a point without a circle, two secants be drawn terminating in the concave arc, they will be reciprocally proportional to their external segments. Let OB and OC be two secants terminating in the concave arc of the circle BCD: then will OB : OC :: OD : OA. For, draw AC and DB. The triangles ODB and OAC have the angle O common, and the angles OBD and OCA equal, because each is measured by half of the arc AD: hence, they are similar, and consequently, their homologous sides are proportional; whence, OB : OC :: : OD : OA ; that is, the rectangles of each secant and its external segment are equal. If from a point without a circle, a tangent and a secant be drawn, the secant terminating in the concave arc, the tangent will be a mean proportional between the secant and its external segment. Let ADC be a circle, OC a secant, and OA a tangent then will OC : OA :: OA : OD. For, draw AD and AC. The triangles OAD and OAC will have the angle O common, and the angles OAD and ACD equal, because each is measured by half of the arc AD (B. III., P. XVIII., P. XXI.); the triangles are therefore similar, and consequently, their A D |