5. The sum of the squares on the sides of a parallelogram is equal to the sum of the squares on its diagonals. PROP. XI.—Problem. To divide a given finite line (AB) into two segments (in H), so that the rectangle (AB.BH) contained by the whole line and one segment may be equal to the square on the other segment. Sol.—On AB describe the square ABDC [I. xlvi.]. Bisect AC in E. Join BE. Produce EA to F, and make EF equal to EB. On AF describe the square AFGH. H is the point required. Dem.—Produce GH to K. Then because CA is bisected in E, and divided externally in F, the rectangle CF.AF, together with the square on EA, is equal to the square on EF [vi.]; but EF is equal to EB (const.); therefore the rectangle CF.AF, together with EA2, is equal to EB2—that is [I. xlvii.] equal to EA2 + AB2. Rejecting EA2, which is common, we get the rectangle CF.AF equal to AB2. Again, since AF is equal to FG, being the sides of a square, the rectangle CF.AF is equal to CF.FG—that is, to the figure CG; and AB2 is equal to the figure AD; therefore CG is equal to AD. Reject the part AK, which is common, and we get the figure FH equal to the figure HD; but HD is equal to the rectangle AB.BH, because BD is equal to AB, and FH is the square on AH. Therefore the rectangle AB.BH is equal to the square on AH. Def.—A line divided as in this Proposition is said to be divided in “extreme and mean ratio.” Cor. 1.—The line CF is divided in “extreme and mean ratio” at A. Cor. 2.—If from the greater segment CA of CF we take a segment equal to AF, it is evident that CA will be divided into parts respectively equal to AH, HB. Hence, if a line be divided in extreme and mean ratio, the greater segment will be cut in the same manner by taking on it a part equal to the less; and the less will be similarly divided by taking on it a part equal to the difference, and so on, &c. Cor. 3.—Let AB be divided in “extreme and mean ratio” in C, then it is evident (Cor. 2) that AC is greater than CB. Cut off CD = CB; then (Cor. 2) AC is cut in “extreme and mean ratio” at D, and CD is greater than AD. Next, cut off DE equal to AD, and in the same manner we have DE greater than EC, and so on. Now since CD is greater than AD, it is evident that CD is not a common measure of AC and CB, and therefore not a common measure of AB and AC. In like manner AD is not a common measure of AC and CD, and therefore not a common measure of AB and AC. Hence, no matter how far we proceed we cannot arrive at any remainder which will be a common measure of AB and AC. Hence, the parts of a line divided in “extreme and mean ratio” are incommensurable. Exercises. 1. Cut a line externally in “extreme and mean ratio.” 2. The difference between the squares on the segments of a line divided in “extreme and mean ratio” is equal to their rectangle.