(« + b)2 + (a−b)2 + 2(a + b)( a − b) = (2a)2, where a and b denote segments. 5. If a given segment be divided into any three parts the square on the segment is equal to the sum of the squares on the parts together with twice the sum of the rectangles on the parts taken two and two. 6. Prove, by comparison of areas from the Fig. of Ex. 1, that (a+b)2=2b(a+b)+2b(a−b)+(a - b)2, and state the theorem in words. SECTION IV. AREAL RELATIONS. 167°. Def.-1. The segment which joins two given points is called the join of the points; and where no reference is made to length the join of two points may be taken to mean the line determined by the points. 2. The foot of the perpendicular from a given point to a given line is the orthogonal projection, or simply the projection, of the point upon the line. 3. Length being considered, the join of the projection of two points is the projection of the join of the points. P Thus if L be a given line and P, Q, two given points, and PP', QQ' perpendiculars upon L; PQ is the join of P and Q, P' and Q' are the projections of P and Q upon L, and the segment P'Q' is the projection of PQ upon L. P Q L 168°. Theorem.-The sum of the projections of the sides of any closed rectilinear figure, taken in cyclic order with AA', BB', CC', DD', and the sum of the projections becomes A'B' + B'C'+C'D'+D'A'. But D'A' is equal in length to the sum of the three others and is opposite in sense. .. the sum is zero. It is readily seen that since we return in every case to the point from which we start the theorem is true whatever be the number or disposition of the sides. This theorem is of great importance in many investigations. Cor. Any side of a closed rectilinear figure is equal to the sum of the projections of the remaining sides, taken in cyclic order, upon the line of that side. Def. In a right-angled triangle the side opposite the right angle is called the hypothenuse, as distinguished from the remaining two sides. A 169°. Theorem.-In any right-angled triangle the square on one of the sides is equal to the rectangle on the hypothenuse and the projection of that side on the hypothenuse. ABC is right-angled at B, and BD is HAC. Then AB2-AC. AD. - Proof. Let AF be the☐ on AC, and let EH be || to AB, and AGHB be a, since LB is a LGAB=LEAC=& LCAB LEAG. (82°, Cor. 5) (hyp.) (64°) AG=AB, and AH is the on AB. Now i.e., AHABLE=ADKE, (140°) q.e.d. As this theorem is very important we give an alternative proof of it. Proof.-AF is the on AC and AH is the on AB, and BD is AC. Cor. 1. Since AB2=AC. AD we have from symmetry ... adding, or BC2=AC. DC, AB2+ BC2 AC(AD+DC), :. The square on the hypothenuse of a right-angled triangle is equal to the sum of the squares on the remaining sides. This theorem, which is one of the most important in the whole of Geometry, is said to have been discovered by Pythagoras about 540 B.C. Cor. 2. Denote the sides by a and c and the hypothenuse by b, and let a1 and c1 denote the projections of the sides a and cupon the hypothenuse. Cor. 3. Denote the altitude to the hypothenuse by p. i.e., the square on the altitude to the hypothenuse is equal to the rectangle on the projections of the sides on the hypothenuse. Def. The side of the square equal in area to a given rectangle is called the mean proportional or the geometric mean between the sides of the rectangle. Thus the altitude to the hypothenuse of a right-angled ▲ is a geometric mean between the segments into which the altitude divides the hypothenuse. (169°, Cor. 3) And any side of the ▲ is a geometric mean between the hypothenuse and its projection on the hypothenuse. (169°) 170°. Theorem.—If the square on one side of a triangle is equal to the sum of the squares on the remaining sides, the triangle is right-angled at that vertex which is opposite the side having the greatest square. (Converse of 169°, Cor.) If AC2=AB2+ BC2, the LB is a B 171o. Theorem 169° with its corollaries and theorem 170° are extensively employed in the practical applications of Geometry. If we take the three numbers 3, 4, and 5, we have 52=32+42. Therefore if a triangle has its sides 3, 4, and 5 feet, metres, miles, or any other u.., it is right-angled opposite the side 5. For the segments into which the altitude divides the hypothenuse we have 51=32 and 5=42, whence a1= and 4. For the altitude itself, p2=.1; whence p=1. Problem. To find sets of whole numbers which represent the sides of right-angled triangles. This problem is solved by any three numbers x, y, and ≈, which satisfy the condition x2= y2+z2. Let m and n denote any two numbers. Then, since (m2 + n2)2 = (m2 — n2)2+(2mn)3, (166°, 4) the problem will be satisfied by the numbers denoted by m2 + n2, m2 — n2, and 2mn. The accompanying table, which may be extended at pleasure, gives a number of sets of such numbers : |