NUMERICAL RELATIONS BETWEEN THE DIFFERENT PARTS OF A TRIANGLE. 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 the I from the rt. _A to the hypotenuse BC. (1) To prove the As DAB, DAC, and ABC similar. B (309) Also, in the rt. As DAC, ABC, the acute ZC 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. Q.E.D. (2) To prove BD: AD = AD: CD. .:. BD: AD = AD: CD. (3) To prove BC : AB= AB: BD. (307) .:. BC : AB = AB: BD. (30%) Also, since the As CAB, CAD are similar, .: BC: AC= AC: CD. 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, AC = BC X CD. AB BD 2 2 A 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 B D in the circumference of a circle to a diamter 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. Der. 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 C 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. A B To prove = D 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. A, with rt. BC = AB’ + AC'. (324) and AC = BC X CD. . (324) Adding, AB' + AC' = BC(BD + CD) = BO'. 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. L, the hypotenuse a is given by the formula a = b + c'. ..a = V0 +0. Thus, if b = 3, c = 4, we have a = 19 + 16 = V25 = 5. If the hypotenuse a and one of the sides b of the rt. Z be known, the other side c is found by the formula C = a* - 7. ..c= Va - 6. ’ Thus, for a = 5, and b = 3, we have c= V25 9 16 = 4. 329. Sch. 2. If AC is the diagonal of a square ABCD, we have (327) A АБ Thus, the diagonal and side of a square are two incommensurable lines, since their ratio, the square root of 2, is an incommensurable number. 2 AC = 12. = 2 NOTE.—To simplify the enunciations of the following theorems, the term square of a line will be given to the second power of the number wbich expresses the measure of the line. The term product of two lines will be given to the product of two numbers which express the magnitude of those lines, measured with the same unit. Proposition 26. Theorem. 330. 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 diminished by twice the product of one of these sides by the projection of the other side 2 = AD + BD + BC-2 BC XBD, by Algebra*, = AB' + BO – 2BC X BD. (327) Q.E.D. * The square of the difference of two numbers is equal to the sum of the squares of the two numbers diminished by twice their product. Proposition 27. Theorem. 331. In an obtuse-angled triangle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the other two sides increased by twice the product of one of these sides by the projection of the other side upon it. A 2 2 2 B Hyp. Let B be the obtuse 2 of the A ABC, and BD the projection of AB upon CB produced. To prove AC` = AB’ + BC+ 2BC X BD. Proof. CD = CB + BD. ... CD' = CB* + BD' + 2BC X BD. (Algebra*) ' Add AD to both sides. .. AC AB + BO' + 2BC X BD. (327) Q.E.D. 332. Cor. From the three preceding theorems, it follows that the square of the side of a triangle is less than, equal to, or greater than, the sum of the squares of the other two sides, according as the angle opposite this side is acute, right, or obtuse. NOTE.–By means of these three theorems we may compute the altitudes of a triangle when the three sides are known. EXERCISES. 1. If the side BC of an equilateral triangle ABC be produced to D, prove that the square on AD is equal to the squares on DC and CA with the product of DC and CA. 2. If ABC be an isosceles triangle having AB = AC, and if D be taken in AC so that BD = BC, prove that the square on BC = AC X CD. * The square of the sum of two numbers is equal to the sum of the squares of the two numbers increased by twice their product. |