ALGEBRAICAL ILLUSTRATION. Let AB = a, CD = b, CE = c, EF = d, FD = e ; then b = c + d + e. Now ABCD = ab, and AB.CE + AB. EF + AB. FD = ac + ad + ae. But since b = c + d + e, .. ab = ac + ad + ae; .. ABCD = AB • CE + AB. EF + AB. FD. 1. The rectangle contained by two straight lines is equal to twice the rectangle contained by one of them and half of the other. 2. The rectangle contained by two straight lines is equal to thrice the rectangle contained by one of them and one-third of the other. 3. The rectangle contained by two equal straight lines is equal to the square on either of them. 4. If two straight lines be each of them divided internally into any number of segments, the rectangle contained by the two straight lines is equal to the several rectangles contained by all the segments of the one taken separately with all the segments of the other. PROPOSITION 2. THEOREM. If a straight line be divided internally into any two segments, the square on the straight line is equal to the sum of the rectangles contained by the straight line and the two segments. A C B D F E Let AB be divided internally into any two segments AC, CB: it is required to prove AB2 = AB.AC + AВ. СВ. On AB describe the square ADEB, and through C draw CF || AD, meeting DE at F. Then AE = AF + CE; that is, AB2 = DA • AC + EB·CB. But DA and EB are each = AB; .`. AB2 = AB. AC + AB • СВ. I. 46 I. 31 I. Ax. 8 ALGEBRAICAL ILLUSTRATION. Let AC = a, CB = b; then AB = a + b. Now, AB2 = (a + b)2 = a2 + 2ab + b2, and ABAC + AB • CB = (a + b) a + (a + b) b = a2 + 2ab + b2; .. AB2 = AB.AC + AB. СВ. 1. Prove this proposition by taking another straight line = AB, and using the preceding proposition. 2. If a straight line be divided internally into any three segments, the square on the straight line is equal to the sum of the rectangles contained by the straight line and the three segments. 3. If a straight line be divided internally into any number of segments, the square on the straight line is equal to the sum of the rectangles contained by the straight line and the several segments. Show that the proposition is equivalent to either of the following: 4. The square on the sum of two straight lines is equal to the two rectangles contained by the sum and each of the straight lines. 5. The square on the greater of two straight lines is equal to the rectangle contained by the two straight lines together with the rectangle contained by the greater and the difference between the two. PROPOSITION 3. THEOREM. If a straight line be divided externally into any two segments, the square on the straight line is equal to the difference of the rectangles contained by the straight line and the two segments. A D B 0 Let AB be divided externally into any two segments АС, СВ: it is required to prove AB2 = AB.AC – AB • СВ. On AB describe the square ADEB, and through C draw CF || AD, meeting DE produced at F. Then AE = AF – CE; that is, AB2 = DA • AC – ЕВ. СВ. But DA and EB are each = AB; .. AB2 = AB. AC – AB. СВ. I. 46 I. 31 I. Ax. 8 NOTE. The enunciation of this proposition usually given is : If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts is equal to the rectangle contained by the two parts together with the square on the aforesaid part. That is, in reference to the figure, AC • AB = AB2 + AB • BC, an expression which can be easily derived from that in the text. ALGEBRAICAL ILLUSTRATION. Let AC = a, CB = b; then AB = a - b. Now, AB2 = (a – b)2 = a2 – 2ab + b2, and ABAC – AB·CB = (a – b) a – (a – b) b = a2 – 2ab + b2; ... AB = AB.AC – AB.CB. 1. Prove this proposition by taking another straight line = AB, and using the first proposition. Show that the proposition is equivalent to either of the following: 2. The rectangle contained by the sum of two straight lines and one of them is equal to the square on that one together with the rectangle contained by the two straight lines. 3. The rectangle contained by two straight lines is equal to the square on the less together with the rectangle contained by the less and the difference of the two straight lines. PROPOSITION 4. THEOREM. If a straight line be divided internally into any two segments, the square on the straight line is equal to the squares on the two segments increased by twice the rectangle contained by the segments. Let AB be divided internally into any two segments AC, CB: it is required to prove AB2 = AC2 + CB2 + 2 AC. CB. On AB describe the square ADEB, and join BD. I. 46 Through C draw CF || AD, meeting DB at G; and through G draw HK || AB, meeting DA and EB at H and K. Because CG || AD, I. 31 I. 5 and because AD = AB, ... LADB = L ABD; .. LCGB = ∠ ABD, = L CBG; CB = CG. I. 6 .. Hence the || CK, having two adjacent sides equal, has all its sides equal. I. 34 But the || CK has one of its angles, KBC, right, = HF + CK + AG + GE, I. Ax. 8 = AC2 + CB2 + 2 AC• СВ. COR. 1. The square on the sum of two straight lines is equal to the sum of the squares on the two straight lines, increased by twice the rectangle contained by the two straight lines. For if AC and CB be the two straight lines, then their sum = AC + CB = AB. Now since AB2 = AC2 + CB2 + 2 AC• СВ, ... (AC + CB)2 = AC2 + CB2 + 2 AC. CB. 2 II. 4 COR. 2.-The ||ms about a diagonal of a square are themselves squares. [It is recommended that II. 7 be read immediately after II. 4.] = (AC.AC + BC·AC) + (AC• BC + BC • BC), II. 2 II. 3 |