Proposition 22. Theorem. 321. Conversely, two similar polygons may be divided into the same number of triangles, similar each to each, and similarly placed. Hyp. Let ABCDE, A'B'C'D'E' be two similar polygons divided into As by the diagonals AC, AD, A'C', A'D' drawn from the homologous s A and A'. To prove As ABC, ACD, ADE sim ilar respectively to As A'B'C', A'C'D', A'D'E'. Proof. Since the polygons are similar, BC A E = (307) As In the same way it may be shown that AS ADE, A'D'E' are similar. Q.E.D. COR. The homologous diagonals of two similar polygons are proportional to the homologous sides. Proposition 23. Theorem. 322. The perimeters of two similar polygons are to each other as any two homologous sides. Hyp. Let ABCDE, A'B'C'D'E' be two similar polygons; denote their perimeters by P and P'. To prove P: P′ = AB : A'B'. Proof. Since the polygons are similar, E B CD P = etc. (307) AB (296) ... P: P' = AB : A'B'. Q.E.D. AB+ BC+ CD + etc. EXERCISES, 1. From the ends of a side of a triangle any two straight lines are drawn to meet the other sides in P, Q; also from the same ends two lines parallel to the former are drawn to meet the sides produced in P', Q': show that PQ is parallel to P'Q'. Let ABC be the ; BP, CQ the lines. AC: AP: = AQ' AB, etc. 2. ABC is a triangle, AD any line drawn from A to a point D in BC; a line is drawn from B bisecting AD in E and cutting AC in F: prove that BF is to BE as 2 CF is to AC. Draw EG || to AC meeting BC in G. DE EA, .. AC 2 EG, etc. 3. If in Prop. 20, AD = 24, AD' = 20, AE = 16, BD = 8, and DE 6, find B'D', D'E', and AE'. NUMERICAL RELATIONS BETWEEN THE DIFFERENT Proposition 24. Theorem. 323. In a right triangle, if a perpendicular be drawn from the right angle to the hypotenuse: (1) The two triangles on each side of it are similar to the whole triangle and to each other. (2) The perpendicular is a mean proportional between the segments of the hypotenuse. (3) Each side about the right angle is a mean proportional, between the hypotenuse and the adjacent segment. Hyp. Let ABC be a rt. A, and AD thefrom the rt. A to the hypotenuse BC. (1) To prove the As DAB, DAC, and ABC similar. Proof. In the rt. As DAB, ACB, the acute / B is common. B D .. the As DAB, ACB are similar. (309) Also, in the rt. As DAC, ABC, the acute C is common. .. the As DAC, ABC are similar. (309) .. the As DAB, DAC are similar to each other, as they are both similar to ABC. Proof. Since the As BAC, BAD are similar, Q.E.D. (307) .. BC: AB = AB : BD. Also, since the ▲s CAB, CAD are similar, .. BC: AC = AC: CD. (307) Q. E. D. 324. COR. 1. If the lines of the figure are expressed in numbers by means of their numerical measures (277), we have from the above three proportions, by means of (281), AD = BD x CD, Hence, the squares of the sides about the right angle are proportional to the adjacent segments of the hypotenuse. 325. COR. 2. Since an angle inscribed in a semicircle is a right angle (240), therefore: (323) (1) The perpendicular from any point in the circumference of a circle to a diam B ter is a mean proportional between the segments of the diameter. (2) The chord from the point to either extremity of the diameter is a mean proportional between the diameter and the adjacent segment. 326. DEF. The projection of a point A upon an indefinite straight line XY is the foot C of the perpendicular let fall from the point to the line. X A The projection of a finite straight line AB upon the line XY is the part of XY between the perpendiculars dropped from the ends of the line AB. Thus, CD is the projection of AB upon the line XY. Proposition 25. Theorem. 327. The square of the number which measures the hypotenuse of a right triangle is equal to the sum of the squares of the numbers which measure the other two sides. Hyp. Let ABC be a rt. ▲, with rt. Adding, AB2 + AC2 = BC(BD + CD) = BC3. Q.E.D. 328. SCH. 1. By this theorem one of the sides of a right triangle can be found when the other sides are known. If we know the two sides b and c about the rt. 2, the hypotenuse a is given by the formula Thus, if b 3, c = 4, we have a = √9 + 16 = √25 = 5. If the hypotenuse a and one of the sides b of the rt. be known, the other side c is found by the formula 329. SCH. 2. If AC is the diagonal of a square ABCD, we have 2 (327) AC2 = AB2 + BC2, or AC2 = 2 AB'. Thus, the diagonal and side of a square are two incommensurable lines, since their ratio, the square root of 2, is an incommensurable number. |