On A B describe the ABDE. Draw the diagonal A D. Through C draw CF parallel to A E or BD, intersecting AD in G and meeting E D in F. Through G draw KGH parallel to It is self-evident that the whole AD is equal to the sum of the four rectangles KC, G B, E G and FH, for they are parts into which it is divided. But KC and F H are □s (Prop. A.), and KC is the on A C, and F H is the ☐ on CB, for it is the ☐ on GH which is equal to C B. (I. 34.) E G is the rectangle contained by A C and C B, for it is contained by K G, which is equal to A C (I. 34) and F G, which is equal to F D, which is equal to C B. (I. 34.) GB is also the rectangle contained by A C and CB, for it is contained by G C and C B, and G C and A C are equal because they are sides of the same square. Consequently EG and GB are together equal to twice the rectangle* contained by AC and C B. We have thus shown that the on the whole line (AB) is equal to the sum of the s on the two parts (A C and CB) together with twice the rectangle under the two parts. The beginner will observe that the stress of the proof does not consist in showing that the whole figure AD is * It would do equally well to point out that the rectangles E G and G B are equal, being complements of the parallelograms K C and F H about the diagonal, and that G B is the rectangle contained by AC and CB; so that EG and G B together are equal to twice the rectangle contained by A C and CB. equal to the sum of the parts into which it is divided, for that is self-evident, but in showing that the whole figure and its parts answer to the description given of them in the enunciation. PROPOSITION V. If a right line be divided into two equal parts, and also into two unequal parts, 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. For the construction employed we must be able— 1. On a given straight line to describe a square. (I. 46.) 2. Through a given point to draw a line parallel to a given right line. (I. 31.) We must also know, 1. That if equals are added to equals, the sums are equal. (Ax. II.) 2. That the opposite sides of a parallelogram are 3. That the parallelograms about the diagonal of 5. That parallelograms upon equal bases and E B Let the line AB be divided into two equal parts in C, and into two Lunequal parts in D. We have to prove that rect. A D, DB, together with the on CD, is equal to the on CB. On C B describe the CBE F. Draw the diagonal FB.* Through D draw the right line D G parallel to CF or B E, intersecting FB in H, and FE in G. Through H draw L K parallel to B A, and through A draw A K parallel to C F or BE, and meeting K L in the point K. 1. Because A C and C B are equal, A O and CL are parallelograms standing on equal bases and between the same parallels; therefore the rectangle A O is equal to the rectangle C L. 2. The rectangles CH and HE are equal, because they are the complements of the parallelograms O G and D L, which are about the diagonal F B. 3. If the equal rectangles CH and H E be added respectively to the equal rectangles A O and CL, the sums will be equal; that is to say, the rectangle A H (which is the sum of AO and C H) will be equal to the gnomon CLG (which is the sum of CL and G L). 4. If to each of these equals the OG be added, the sums will be equal; that is to say, the sum of the rect. A H and the OG will be equal to the sum of the gnomon CL G and the O G. 5. But A H is the rectangle contained by A D and DB, for it is contained by A D and D H, and DH and DB are equal, being sides of the same; and O G on CD, for it is the □ on OH which is equal to CD. (I. 34.) is the 6. Therefore the rect. AD, DB together with the on CD is equal to the gnomon CL G together with theO G. * Let the beginner take particular notice that the diagonal needed is the one which is drawn to the extremity of the line AB. The diagonal C E would not do. L 7. But the gnomon CLG together with the OG is equal to the CE, which is the on CB. Therefore the rectangle under the unequal parts, A D and D B, together with the □ on C D, the part between the points of section, is equal to the ☐ on CB, which is half the line A B. The steps of this proof may be easily indicated by the use of the following equations, which express what is stated in the preceding paragraphs which bear the same numbers : 1. Rect. A O= rect. CL. 2. Rect. CH=rect. HE. 3. Therefore rect. AO+rect. CH= rect. CL+ rect. HE, or rect. A H = gnomon CL G. 4. Add to both the OG gnomon CLG+ OG, then rect. A H + O 0 G. 5. But A H = rect. A D, D B, and O G = on CD. . 6. Therefore rect. A D, D B +□ on CD = gnomon CLG+□OG. 7. But gnomon CLG+ OG = on CB. 8. Therefore rect. A D, D B + □ on CD = ☐ on CB. If AC and CD be regarded as two separate lines, it is clear that AD is the sum of the two, and D B the difference between them, for D B is what is left when CD is taken from C B, and C B is equal to A C. Therefore A H is the rectangle under the sum and the difference of the two lines A C and CD. OG is the on the smaller of the two lines; and CE is the ☐ on the larger, for it is the on CB, which is equal to A C. Consequently what is proved in this proposition may be thus stated: "The rectangle contained by the sum and the difference of two unequal lines, together with the on the smaller, is equal to the "The difference between the on the larger;" or thus: s on two unequal lines (as AC and CD) is equal to the rectangle contained by their sum and their difference.*" This is incomparably the most useful form in which the proposition can be stated. PROPOSITION VI. If a right line be bisected and produced to any length, the rectangle under the whole line thus produced and the produced part, together with the square on half the line, is equal to the square on the line made up of the half line and the produced part. The propositions required for the construction and proof in this proposition are the same as in the last. A C B D Let A B be a right line bisected in C, and produced to D. We have to prove that the rect. A D, DB, together with the on CB, is equal to the☐ on CD. On CD describe the CDEF. Draw the diagonal F D (see the note on the last proposition.) Through B draw BG parallel to CF or D E, cutting FD in H and FE in G. Through H draw KHM parallel to AD, and through A draw AM parallel to D E, and meeting KM in the point M. 1. The rectangles A L and CH are on equal bases *The beginner will remember that the difference between two magnitudes is what is added to the smaller to make a quantity equal to the larger; and the rectangle AH is added to the O G to make a quantity equal to the CE. |