(A C and C B) and between the same parallels, therefore they are equal to each other. 2. The rectangles C H and H E are the complements of the parallelograms L G and B K, which are about the diagonal FD; therefore they are equal to each other. 3. It follows therefore that the rect. A L is equal to the rect. H E. 4. If to each of these equal rectangles we add the rect. C K, the sums will be equal, that is, the rect. AK will be equal to the gnomon C KG. 5. If now to each of these equals we add the LG, the rect. A K together with the L G will be equal to the gnomon CKG together with the LG, that is, to the whole o CE. 6. But A K is the rectangle contained by A D and D B, for it is contained by AD and D K, and D K and D B are equal, being sides of the same square. 7. LG is the oon CB, for it is the on LH, and LH is equal to CB (I. 34); and CE is the on CD. 8. We have thus shown that the rectangle contained by the whole produced line (A D) and the part produced (BD), together with the on half the line (CB), is equal to the on CD, which is the line made up of the half CB and the produced part BD. The steps of the preceding proof are exhibited in tho following equations : 1. Rect. AL= rect. C H. Rect. AL + rect. CK= rect. HE + rect. C K. Or rect. A K=gnomon C KG. ő. Add to both the LG, then Rect. A K +OLG=gnomon CKG+OLG. or rect. A K+OLG=0CE. 6. But rect. A K= rect. A D, D B, and 7. OLG= 0 on C B, and a CE is a on C D. 8. Therefore rect. AD, DB+ on CB= 0 on CD. If we regard A C and C D as two separate lines, it is clear that A D is their sum, and BD their difference, for BD is what is left when C B (or A C, which is equal to CB) is taken from CD. CE is the on CD, and LG is the on A C, for it the on L H, which is equal to C B which is equal to A C. The difference between the Os CE and L G is the gnomon C KG, which is equal to the rectangle A K. Consequently this proposition may be expressed in exactly the same form as the last, namely: The difference between the Os on two lines (CD and C A) is the rectangle contained by their sum (A D) and their difference (B D.) The segments of a right line made by a point in it are the distances of the point from the two extremities of the line. This definition of segments may be extended to the case where the point is not in the line itself, but in the line produced ; so that in Prop. VI. D A and D B are termed segments of the line A B. The fifth and sixth propositions may now be combined into one in this way: The rectangle contained by the segments of a line formed by a point (either in the line or in the line produced) is the difference between the on half the line and the on the distance of the point of section from the middle of the line. In the fifth proposition the former of these two s is the larger, in the sixth, the latter. PROPOSITION VII. If a straight line be divided into any two parts, the squares on the whole line and on one of the parts are together equal to twice the rectangle contained by the whole and that part together with the square on the other part. E F K A B The propositions required for the construction and proof in this proposition are the same as in the fourth, together with Prop. A (p. 140.) Let A B be a straight line divided into two parts at the point C. We have to show that the con AB together with the on one part (say C B) is equal to twice the rect. A B, B C together with the on A C. Through C draw CF parallel to A E or BD, intersecting E B in H, and E D in F. Through H draw GIK parallel to A B, meeting BD in G and A Ein K. 1. KF and C G are Os (Prop. A): and KF is the o on A C, for it is the ou KH, and K H is equal to A C (I. 34); and CG is the on CB. 2. It is self-evident that the A D is equal to the rectangles A G and HD, together with the KF. 3. To each of these equals add the CG, and the Os A D and CG together will be equal to the sum of the OKF, the rect. A G, the rect. HD, and the o CG. 4. But the rect. HD together with the o C G is equal to the rect. C D. 5. Therefore the D A D together with the OCG is equal to the KF, the rect. A G and the rect. CD taken together. 6. But the rectangles A H and HD are equal, being complements of the parallelograms about the diagonal of A D. Therefore the sum of AH and CG (i.e., the rectangle A G) is equal to the sum of HD and CG (i.e., the rectangle CD.) 7. Therefore the sum of A G and C D is the same as twice the rectangle A G, which is the rectangle under A B and B C, for BG is equal to B C. 8. It follows therefore that the on A B, together with the on C B is equal to twice the rect. A B, BC, together with the KF, which is the on A C. The beginner must not fail to notice that C G is the piece by which the rectangles A G and CD overlap each other. The leading steps of the above demonstration may be clearly shown by the aid of equations : 2. O AD OKF + rect. A G, + rect. HD. 3. Add 0 CG to both sides; then O AD + OCG =KF + rect. A G + rect. HD + O CG. 4. But rect. HD + O CG= rect. C D. 5. Therefore AD + OCG= 0 KF + rect. AG + rect. CD. 6. Now rect. AH= rect. HD. Therefore rect. A H + O CG = rect. HD + OCG, that is, rect. A G = rect. CD. 7. Therefore rect. AG + rect. CD= 2c rect. AG, or rect. A G + rect. CD= 2c rect. A B, BC. 8. O AD + OCG= OKF + 2c rect. A G, or on AB + Oon BC= 2c rect. A B, BC + on A C. In this proposition again the learner will observe that the bulk of the proof consists in showing that the various squares and rectangles compared with each other answer to the description given of them in the enunciation. Prop. VIII. is usually passed over, as it is troublesome and useless. PROPOSITION LX. If a straight line be divided into two equal and also into two unequal parts, the squares on the two unequal parts are together equal to twice the square on half the line together with twice the square on the line between the points of section. same. In the construction employed in this proposition we must be able,1. To join two given points by a straight line. (Post. I.) 2. From the greater of two given straight lines to cut off a part equal to the less. 3. To draw a straight line perpendicular to a given straight line from a given point in the (I. 11.) 4. Through a given point to draw a straight line parallel to a given straight line. (I. 31.) To prove the proposition by the aid of the figure constructed, we must know,1. That the <s at the base of an isosceles tri angle are equal. (I. 5.) 2. That if two Zs of a triangle are equal, the sides opposite them are also equal. (I. 6.) 3. That the three xs of a triangle are together equal to two right Zs. (I. 32.) 4. That if a straight line intersect two parallel straight lines, the exterior _ is equal to the interior opposite _ on the same side of the intersecting line, and the two interior <8 on the same side are together equal to two right 8. (I. 29.) 5. That the opposite sides of a parallelogram are equal. (I. 34.) |