III. The discussion is left as an exercise for the student. 441. The perimeters of two similar polygons are to each other as any two homologous sides. Given ~ polygons P and Q, with sides a, b, c, d, e, and f homol. respectively to sides k, l, m, n, o, and p. Ex. 699. The perimeters of two similar polygons are 152 and 138; a side of the first is 8. Find the homologous side of the second. Ex. 700. The perimeters of two similar polygons are to each other as any two homologous diagonals. Ex. 701. The perimeters of two similar triangles are to each other as any two homologous medians. Ex. 702. If perpendiculars are drawn to the hypotenuse of a right triangle at its extremities, and if the other two sides of the triangle are prolonged to meet these perpendiculars, the figure thus formed contains five triangles each of which is similar to any one of the others. PROPOSITION XXV. THEOREM 442. In a right triangle, if the altitude upon the hypotenuse is drawn: I. The triangles thus formed are similar to the given triangle and to each other. II. The altitude is a mean proportional between the segments of the hypotenuse. III. Either side is a mean proportional between the whole hypotenuse and the segment of the hypotenuse adjacent to that side. Given rt. ▲ ABC and the altitude CD upon the hypotenuse. To prove : I. ▲ BCD ~ ^ ABC, ▲ ADC ~ ▲ ABC, and ▲ BCD ~ ▲ ADC. II. BD: CD = CD: DA. III. AB: BC= BC: DB and AB: AC= AC: AD. II. The proof of II is left as an exercise for the student. HINT. Mark (with colored chalk, if convenient) the lines required in the proportion. Decide which triangles will furnish these lines, and use the fact that homologous sides of similar triangles are proportional. III. The proof of III is left as an exercise for the student. HINT. Use the same method as for II. 443. Cor. I. In a right triangle, if the altitude upon the hypotenuse is drawn: I. The square of the altitude is equal to the product of the segments of the hypotenuse. Thus, CD BD. DA. (See § 442, II.) = II. The square of either side is equal to the product of the whole hypotenuse and the segment of the hypotenuse adjacent to that side. Thus, BC2 = AB · DB, and AC2 = AB AD. (See § 442, III.) = 444. Cor. II. If from any point in the circumference of a circle a perpendicular to a diameter is drawn, and if chords are drawn from the point to the ends of the diameter: I. The perpendicular is a mean proportional between the segments A D of the diameter. B II. Either chord is a mean proportional between the whole diameter and the segment of the diameter adjacent to the chord. HINT. ▲ APB is a rt. ^. Apply Prop. XXV, II and III. Ex. 703. In a right triangle, the squares of the two sides are proportional to the segments of the hypotenuse made by the altitude upon it. HINT. Apply § 443, II. Ex. 704. The sides of a right triangle are 9, 12, and 15. (1) Compute the segments of the hypotenuse made by the altitude upon it. (2) Compute the length of the altitude. Ex. 705. If the two arms of a right triangle are 6 and 8, compute the length of the perpendicular from the vertex of the right angle to the hypotenuse. PROPOSITION XXVI. PROBLEM 445. To construct a mean proportional between two given lines. m n E 1. On any indefinite str. line, as AB, lay off AC=m and CD = n. 2. With 0, the mid-point of AD, as center and with a radius equal to OD, describe a semicircumference. 3. At C construct CEL AD, meeting the semicircumference at E. § 148. 4. CE is the required line 7. II. The proof and discussion are left as an exercise for the student. Ex. 706. By means of § 444, II, construct a mean proportional between two given lines by a method different from that given in Prop. XXVI. Ex. 707. Use the method of Prop. XXVI to construct a line equal to √3 ab, a and b being given lines. ANALYSIS. Ex. 708. Construct a line equal to a√3, where a is a given line. Ex. 709. Using a line one inch long as a unit, construct a line equal to √3; √5; √6. Choosing your own unit, construct a line equal to 3√2, 2√3, 5√5. PROPOSITION XXVII. THEOREM 446. In any right triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides. 447. Cor. I. The square of either side of a right triangle is equal to the square of the hypotenuse minus the square of the other side. 448. Cor. II. The diagonal of a square is equal to its side multiplied by the square root of two. OUTLINE OF PROOF d Ꭶ 1. d2 = s2 + s2 = 2 s2. 2. .. d = s√2. Q.E.D. 8 449. Historical Note. The property of the right triangle stated in Prop. XXVII was known at a very early date, the ancient Egyptians, 2000 B.C., having made a right triangle by stretching around three pegs a cord measured off into 3, 4, and 5 units. See Note, Book IV, § 510. |