4. In any right triangle, the product of the legs equals the product of the hypotenuse and the altitude to the hypotenuse. SUGGESTION. - In § 192, prove AC × BC = AB × CD. This equality would follow from what proportion? The proportion follows from what similar triangles? Hence begin by proving the triangles similar. 5. In any right triangle, the squares of the legs are to each other as their projections upon the hypotenuse. SUGGESTION. - This may be proved by means of the equations in Ex. 3. 6. In any right triangle, the square of the hypotenuse and the square of either leg are proportional to the hypotenuse and the projection of that leg upon the hypotenuse. SUGGESTION. - In the figure of § 192, how is the required proportion derived from the equations AB = AB × AB and AC2 = AB × AD? 194. Theorem. In any right triangle, the altitude to the hypotenuse is the mean proportional between the two segments into which it divides the hypotenuse. Hypothesis. In ∆ABC, ∠ Cis a right angle, and CD LAB. AD CD Conclusion. = CD DB Suggestion. This conclusion will follow if what triangles are proved similar? Write the proof in full. 195. Corollary. - The half of a chord per pendicular to a diameter is the mean proportional between the segments into which it divides the diameter. The proof is left to the student. EXERCISES 1. The altitude to the hypotenuse of a right triangle divides it into segments of 12 in. and 27 in., respectively. Find the altitude to the hypotenuse. 2. The hypotenuse of a right triangle is 26 ft. The altitude to the hypotenuse is 12 ft. Find the segments into which the altitude divides the hypotenuse. 3. If one acute angle of a right triangle is 30°, the altitude to the hypotenuse divides the hypotenuse in the ratio of 1 to 3. 4. If one leg of a right triangle is twice the other, the altitude to the hypotenuse divides it in the ratio of 1 to 4. 5. A railroad surveyor who wished to find the radius of the railroad curve ACB, measured the chord AB and the distance CD from the middle point of the arc to the middle point of the chord. Show how he was then able to compute the radius. If AB = 100 ft. and CD = 3 ft., compute the radius of the curve. 6. A window whose width AB is 5 ft. is to be surmounted by a circular stone arch of which the rise CD is 20 in. Find the radius of the circle at which the stone for the arch must be cut. 7. In a bridge, a circular arch 18 ft. high is to span a stream 72 ft. wide. What is the radius of the circle at which the stones of this arch must be cut? 8. This is a piece of a broken wheel. AB=16 in. and CD=4 in. Find the diameter of the wheel. A C C A D B C B D B 9. The diameter of the earth may be computed as follows: Three stakes are set in a canal two miles long, one at each end and one in the middle, and all project the same distance above the water. By use of a leveling instrument the middle stake is found to be 8 inches higher than the others. From these facts compute the diameter of the earth. 10. If r is the radius of a circle, find the length of a chord whose distance from the center is r. 196. Theorem. In any right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs. Hypothesis. In △ABC, ∠C is a right angle. The legs are a and b, and the hypotenuse c. Conclusion. c2 = a2 + b2. Proof. 1. In △ABC, ∠ C is a right angle. 2. Draw CDI AB, and let AD = m and DB 3. Then = and. C Нур. n. § 192 Ax. IV Ax. II Factoring Ax. XII NOTE. - This theorem is called the Pythagorean Theorem, after the famous Greek philosopher and mathematician Pythagoras (about 540 в.с.), who is believed to have given the first rigorous proof of it. The truth of the theorem for special cases was known by the Egyptians centuries before the time of Pythagoras, and was used by them in building their pyramids. They laid out perpendicular lines by stretching a rope around three pegs so placed that the distances between them were proportional to 3, 4, and 5. The same principle is employed by carpenters and others now. Let the student, if possible, find some builder who is staking off the ground for the foundation of a house, and see how he obtains the right angles at the corners. The proof above is attributed to the Hindus. Another proof of the theorem is given in § 231. It was given in the first great geometry, Euclid's Elements, about 300 в.с. Still other proofs are suggested in the exercises on page 229. Just what proof was given by Pythagoras is not known. This is one of the most important and practical theorems of geometry. 197. Theorem. - In any triangle, the square of the side opposite an acute angle is equal to the sum of the squares of the other two sides minus twice the product of one of those sides and the projection of the other upon it. Hypothesis. sides opposite A, B, and C, respectively, and m the projec In ∆ABC, ∠ A is acute, a, b, and care the Proof. 1. A is acute, a, b, and care the sides opposite A, B, and C, respectively, and m is the projection of b upon c. Нур. 2. Let n be the projection of a upon c, and let altitude CD = h. Then if CD falls within ABC, n = c - т. And if CD falls without △ABC, n = m - с. If CD falls on CB, the triangle is a right triangle. Show that the theorem is true in that case. 198. Theorem. - In any obtuse triangle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the other two sides plus twice the product of one of those sides and the projection of the other upon it. Hypothesis. In △ABC, ∠ A is obtuse, a, b, and c are the sides opposite A, B, and C, respectively, and m the projection of b upon c. Suggestions. Since n = m + c, show that n2=m2+c2+2 ст. Then add ha to each member of the equation, and proceed as in the proof of § 197. Write the proof in full. 199. Theorem. If a, b, and c are the sides opposite A, B, and C, respectively, of △ABC, then: ∠A is acute if a2 < b2 + c2, ∠ A is obtuse if a2 > b2 + c2, ∠A is a right angle if a2 = b2 + c2. Suggestion. Each part of the theorem may be proved indirectly by use of § 196, § 197, and § 198. Thus, for proving the first part, suppose that ∠ A is not acute. Then it is either obtuse or a right angle. If ∠ A is obtuse, a2 > b2 + c2 by § 1983; and if ∠ A is a right angle, a2=b2+c2 by § 196. But these are untrue by hypothesis, etc. Write the proof in full. |