= Again, the complement AG the complement GE; I. 43. = *Now the sq. on AB = the fig. AE = the figs. HF, CK, AG, GE the sqq. on AC, CB together with twice the rect. AC, CB. .. the sq. on AB = the sum of the sqq. on AC, CB with twice the rect. AC, CB. Q.E.D. COROLLARY 1. Parallelograms about the diagonals of a square are themselves squares. COROLLARY 2. If a straight line is bisected, the square on the whole line is four times the square on half the line. *For the purpose of oral work, this step of the proof may conveniently be arranged as follows: Now the sq. on AB is equal to the fig. AE, that is, to the figs. HF, CK, AG, GE; that is, to the sqq. on AC, CB together with twice the rect. AC, CB. CORRESPONDING ALGEBRAICAL FORMULA. The result of this important Proposition may be written thus : hence the statement AB2=AC2+CB2+2AC. CB becomes (a+b)2=a2+b2+2ub. PROPOSITION 5. THEOREM. If a straight line is divided equally and also unequally, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line. A Let the straight line AB be divided equally at P, and unequally at Q. Then the rect. AQ, QB together with the sq. on PQ shall be equal to the sq. on PB. Construction. On PB describe the square PCDB. I. 46. Join BC. Through Q draw QE par1 to BD, cutting BC in F. I. 31. Proof. Now the complement PF the complement FD: = to each add the fig. QL; then the fig. PL = the fig. QD. I. 43. But the fig. PL = the fig. AH, for they are parms on equal bases and between the same parls; .._the fig. AH = the fig. QD. To each add the fig. PF; then the fig. AF = the gnomon PLE. Now the fig. AF is the rect. AQ, QB; for QF=QB ; I. 36. To each add the sq. on PQ, that is, the fig. HE; II. 4. then the rect. AQ, QB with the sq. on PQ = the gnomon PLE with the fig. HE = -the whole fig. PD, which is the sq. on PB. That is, the rect. AQ, QB together with the square on PQ is equal to the sq. on PB. Q.E.D. COROLLARY. From this Proposition it follows that the difference of the squares on two straight lines is equal to the rectangle contained by their sum and difference. From PB cut off PQ equal to Y. Then AQ is equal to the sum of X and Y, and QB is equal to the difference of X and Y. Now because AB is divided equally at P and unequally at Q, .. the rect. AQ, QB with sq. on PQ-the sq. on PB; II. 5. that is, the difference of the sqq. on PB, PQ=the rect. AQ, QB. or, the difference of the sqq. on X and Y the rectangle contained by the sum and the difference of X and Y. In the above figure shew that AP is half the sum of AQ and QB ; and that PQ is half their difference. If a straight line is bisected and produced to any point, the rectangle contained by the whole line thus produced and the part of it produced, together with the square on half the line bisected, is equal to the square on the straight line made up of the half and the part produced. Let the straight line AB be bisected at P, and produced to Q. Then the rect. AQ, QB together with the sq. on PB shall be equal to the sq. on PQ. Construction. On PQ describe the square PCDQ. I. 46. Join QC. Through B draw BE par1 to QD, meeting QC in F. I. 31. Proof. Now the complement PF = the complement FD. I. 43. But the fig. PF= the fig. AH; for they are parms on equal bases and between the same parls. .. the fig. AH = the fig. FD. To each add the fig. PL; then the fig. AL = the gnomon PLE. Now the fig. AL is the rect. AQ, QB; for QL=QB; .. the rect. AQ, QB = the gnomon PLE. To each add the sq. on PB, that is, the fig. HE; then the rect. AQ, QB with the sq. on PB I. 36. the gnomon PLE with the fig. HE - the whole fig. PD, = which is the square on PQ. That is, the rect. AQ, QB together with the sq. on PB is equal to the sq. on PQ. Q.E.D. A X B DEFINITION. If a point X is taken in a straight line AB, or in AB produced, the distances of the point of section from the extremities of AB are said to be the segments into which AB is divided at X. In the former case AB is A divided internally, in the latter case externally. B X Thus in each of the annexed figures, the segments into which AB is divided at X are the lines AX and XB. This definition enables us to include Props. 5 and 6 in a single Enunciation. If a straight line is bisected, and also divided (internally or externally) into two unequal segments, the rectangle contained by the unequal segments is equal to the difference of the squares on half the line, and on the line between the points of section. EXERCISE. Shew that the Enunciations of Props. 5 and 6 may take the following form: The rectangle contained by two straight lines is equal to the difference of the squares on half their sum and on half their difference. [See Ex., p. 137.] |