points of division draw lines ac, bd, &c., parallel to E F. Draw also the lines a e, Bƒ, &c., parallel to DF. By article 91, the triangles A ac, a Be, &c. are equal. By article 92, ae is equal to c C, Bf to C d, &c. Hence it is easy to see that A B is composed of the same number of times A a, that A C is of A c, and that in like manner D E is as many times Da, as D F is of Dc. And thus by article 78, DE: DF: AB: DC, because each of these ratios is equal to Ва: ае. 100. Scholium. = If the lines A B and D E do not consist of a certain number of times the first unit of length which we have chosen, we may choose a unit small enough to make the remainder small enough to be neglected. 101. Definitions. The right angle, right triangle, legs, and hypotenuse, are defined in articles 14 and 17. 102. Theorem. The sum of the three angles of a triangle is equivalent to two right angles. This proposition has been proved in articles 26–31, 34-36, and 57–62. 103. Corollary. The sum of the two angles opposite to the legs of a right triangle is equivalent to one right angle. 104. Corollary. If an angle opposite a leg in one right triangle is equal to an angle in another right triangle, the two right triangles are equiangular with respect to each other. 105. Theorem. If from the vertex of the right angle in a right triangle, a line be drawn at A B' Fig. A. E F G B right angles to the hypotenuse dividing the hypotenuse into two segments, each leg is a mean proportional between the whole hypotenuse and the segment nearest the leg.-Proof. Let A B C be a right triangle with a right angle at C. Draw CF at right angles to A B. The triangle BEC is right angled at E, and has an angle at B equal that at B in the triangle A B C. Hence, by article 104, the triangle BEC has its angles equal to those of A B C. Hence, by article 99, B E: B C = B C : BA. In the same way AE: AC:: AC: A B. 106. Theorem. The square on the hypotenuse is equivalent to the sum of the squares on the legs. —Proof. Let A CB (Fig. A.) be a right triangle, with a right angle at C, and let a square be drawn on each side. Draw CF at right angles to A B. The figure B F will be a rectangle, because all its angles will be right angles. It will, therefore, be measured by the product of BE into EF; or (since E F BG and B G BA), by the product of BEX BA. But since BC is a mean proportional between the lines BE and B A, this product is equal to BC X B C, which is the measure of the square on B C. That is, the measure of the rectangle B F is the same as that of the square on B C. In the same manner it may be shown that the rectangle AF is equivalent to the square on A C. But the sum of these two rectangles is evidently equal to the square on the hypotenuse. 107. In these thirty-one articles I have given you a proof of the Pythagorean proposition in the usual synthetic form. Parts of the proof are not completely filled out, but the omitted steps are so short and easy, that I think you will have no difficulty whatever in supplying them. Do not be satisfied with understanding each of the thirty-one articles, but examine them closely from the 76th to the 106th, and see whether I have introduced anything which is not necessary to the proof of 106. In making this examination, it will be most convenient for you to proceed backward. These thirty-one articles have been here introduced as lemmas, i. e. preparatory propositions, for demonstrating the Pythagorean proposition. But they are also each one truths worth knowing, and will aid in establishing many theorems that have no connection with a right triangle. 108. Another mode of analyzing this proposition would be suggested by our knowledge of the fact that any triangle is equivalent to half a rectangle of the same base and altitude. I will not lead you through this analysis, but will simply build up for you by synthesis, a Second Proof of the Pythagorean Proposition. 109. Definitions. Any side of a triangle or quadrangle may be called its base, and the altitude of the figure is the distance from the base to the most distant vertex of the figure. This distance is measured by a straight line at right angles to the base, and contained between the vertex and the base, prolonged if need be. Fig. C. C E Ꭰ F 110. Theorem. Every parallelogram is equivalent to a rectangle of the same base and altitude. Proof. Let A B C D be a rectangle, and ABEF a parallelogram having the same base A B, and the same altitude BD. It is manifest that if the triangle BDF by which the parallelogram overlaps the rectangle is equal to the triangle A E C by which the rectangle overlaps the parallelogram, the two quadrangles are equivalent. But A E and its adjacent angles are equal to B F and its adjacent angles, and therefore the triangles are equal (Art. 91), and the quadrangles equivalent. 111. Theorem. Every triangle is equivalent to half a rectangle of the same base and altitude.-Proof. Let A F B be a triangle, and A B C D a rectangle having the same base, A B and the same altitude BD (Fig. C.). Continue CD to F and draw A E parallel to BF. The triangle A EF has its three sides equal to those of A B F; the triangles are, therefore, equal to each other (Art. 95); and each is equal to half the parallelogram ABEF, which is equivalent to the rectangle A B C D. 112. Theorem. The square on the hypotenuse is equivalent to the sum of the squares on the legs.-Proof. Having drawn the figure (Fig. A.), as for the former proof, draw the lines C' B, and B' C. The triangle ABC has the same base A B', and the same altitude A E as the rectangle A F, and is equivalent to half that rectangle. The triangle A B C ́ has the same base A C ́ and the same altitude A C as the square C' C, and is equivalent to half that square. So that if the triangles A B C ́ and A B′ C are equal, the rectangle is equivalent to the square. But these triangles are equal, for if A B' C were turned about the vertex A as on a pivot until the point C covered C', then B' would cover B, and the triangles would coincide. For AC would rotate through a right angle, and A B' through a right angle; and A C― A C', and ABAB'. = 113. This proof of the Pythagorean proposition is more strictly geometrical than the preceding, as it does not involve the idea of multiplying lines to measure areas. But you must remember that each is equally conclusive. I have here also omitted some of the shorter steps. You should not only be able to fill out these steps when the omission is pointed out to you, but also to discover the omission for yourselves. Take the proofs which I have written down and examine them step by step, asking at each step, is that strictly self-evident? Can it be questioned? Can it be divided into two steps? Is there need of proof? If so, has the proof been given in a previous article? It is only by such an earnest study of the book and of the subject that you can make the process of mathematical reasoning become a sure and pleasant road for you to the discovery of truth. [To be continued.] THE following students have sent us solutions of the Prize Problems in the October Number, Vol. II., of the Monthly. MISS AMANDA BENNETT, Red Wing, Minnesota. Prob. I. MISS HARRIET L. ENSIGN, Catskill Academy, N. Y. Probs. I., II., III., and IV. MISS HARRIET S. HAZELTINE, Worcester, Mass. Probs. I., II., and IV. OAKLEY H. SMITH, New Hampton Institution, Fairfax, Vt. Probs. I. and II. D. Y. BINGHAM, Ellicotteville, N. Y. Prob. III. SAMUEL S. EASTWOOD, High School, Saxonville, Mass. Probs. I. and II. C. P. MARSTON, Public High School, Hartford, Ct. Probs. I. and II. WILLIAM W. JOHNSON, Sophomore Class, Yale College, Ct. Probs. III. and IV., and first part of V. BENJAMIN F. WEBBER, Wesleyan Seminary, Kent's Hill, Maine. Probs. I. and II. ISAAC H. TURRELL, Drewersburg, Ind. Probs. I., II., III., and IV. GEORGE D. HALE, Select School, Adams Centre, N. Y. Probs. I. and II. F. E. HASTINGS, Literary Institution, Suffield, Ct. Prob. I. G. W. BROWN and ISAAC FULLER, Collegiate Institution, Bowden, Ga. Each Probs. I., II. WILLIAM C. CLEVELAND, Lawrence Scientific School, Cambridge, Mass. All but V. WILLIAM C. HENCK, High School, Dedham, Mass. Probs. I. and II. CHARLES B. BOUTELLE, Waterville Academy, Maine. Prob. III. M. H. DOOLITTLE, Sophomore Class, Antioch College, Yellow Springs, Ohio. Probs. III., IV., and first part of V. W. D. MERSHON, Sophomore Class, Princeton College, N. J. Prob. IV. Probs. I. and II. R. E. LEONARD, Ward School, No. 32, New York City. Probs. I. and II. and III. J. D. VAN BUREN, Rensselaer Polytechnic Institute, Troy, N. Y. Probs. III., IV., and V. O. B. WHEELER, Sophomore Class, University of Michigan, Ann Arbor. Probs. III., IV., and V. JOSEPH COCKE and AYLETT B. COLEMAN, Locust Grove Academy, Albemarle Co., Va. Each Probs. I. and II. M. K. BOSWORTH, Sophomore Class, Marietta College, Ohio. Probs. III., IV., and V. W. A. AKEN, New Wilmington, Pa. Probs. I. and II. The following solutions, unfortunately, did not reach us in time. WILSON BERRYMAN and ОTHO E. MICHAELIS, Sophomore Class, N. Y. Free Academy, each answered all the questions. H. TIEMAN, Baltimore, Md., answered all the questions but the second part of Prob. V. PHILO HOLCOMB, Hughes High School, Cincinnati, answered Prob. IV. THE MATHEMATICAL MONTHLY. Vol. II. . FEBRUARY, 1860.... No. V. ... PRIZE PROBLEMS FOR STUDENTS. I. Prove that the value of a proper fraction is increased, and an improper fraction diminished, by adding the same quantity to both terms of the fraction; and that the reverse is the case when the same quantity is subtracted from both terms of the fraction. II. A common tangent is drawn to two circles which touch each other externally; if a circle be described on that part of the tangent which lies between the points of tangency as diameter, this circle will pass through the point of contact of the two circles, and will touch the line joining their centres. III. Find the n quantities x1, x2, . . . . . x, from the n equations, Xn and obtain symmetrical expressions for x1, x2, &c. IV. Prove that sin" (0 — 9) sin q is a maximum when V. The notation of Problem V. in the November No. being retained, prove that in the plane triangle REPORT OF THE JUDGES UPON THE SOLUTIONS OF THE PRIZE PROBLEMS IN No. II., Vol. II. THE first Prize is awarded to GEORGE B. HICKS, Cleveland, Ohio. The second Prize is awarded to WILLIAM HINCHCLIFFE, Barre Plains, Mass. The third Prize is not awarded. PRIZE SOLUTION OF PROBLEM III. By GEORGE B. HICKS, Cleveland, Ohio. Of all right-angled plane triangles having the same given hypothenuse, to find the one whose area is the greatest possible. To be solved by Algebra. Since the hypothenuse is constant, the area is evidently a maximum when the perpendicular dropped from the right angle upon the hypothenuse is a maximum. Let x y be the sides, c the hypothenuse, p the perpendicular, and d one of the segments of the hypothenuse. Then |