Now since AE is equal to AB, the angle ABE is equal to AEB [I. v.]; but since BE intersects the parallels AE, CF, the angle AEB is equal to CGB [I. xxix.]. Hence the angle CBG is equal to CGB, and therefore [I. vi.] CG is equal to CB; but CG is equal to BI and CB to GI. Hence the figure CBIG is a lozenge, and the angle CBI is right. Hence (I., Def. xxx.) it is a square. In like manner the figure EFGH is a square. Again, since CB is equal to CG, the rectangle AC.CB is equal to the rectangle AC.CG; but AC.CG is the figure AG (Def. iv.). Therefore the rectangle AC.CB is equal to the figure AG. Now the figures AG, GD are equal [I. xliii.], being the complements about the diagonal of the parallelogram AD. Hence the parallelograms AG, GD are together equal to twice the rectangle AC.CB. Again, the figure HF is the square on HG, and HG is equal to AC. Therefore HF is equal to the square on AC, and CI is the square on CB; but the whole figure AD, which is the square on AB, is the sum of the four figures HF, CI, AG, GD. Therefore the square on AB is equal to the sum of the squares on AC, CB, and twice the rectangle AC.CB. Or thus: On AB describe the square ABDE, and cut off AH, EG, DF each equal to CB. Join CF, FG, GH, HC. Now the four △s ACH, CBF, FDG, GEH are evidently equal; therefore their sum is equal to four times the △ACH; but the △ACH is half the rectangle AC.AH (i. Cor. 2)—that is, equal to half the rectangle AC.CB. Therefore the sum of the four triangles is equal to 2AC.CB. Again, the figure CFGH is a square [I. xlvi., Cor. 3], and equal to AC2 + AH2 [I. xlvii.]—that is, equal to AC2 + CB2. Hence the whole figure ABDE = AC2 + CB2 + 2AC.CB. Or thus: AB = AC + CB. Squaring, we get AB2 =AC2 + 2AC.CB + CB2. Cor. 1.—The parallelograms about the diagonal of a square are squares. Cor. 2.—The square on a line is equal to four times the square on its half. For let AB = 2AC, then AB2 = 4AC2. This Cor. may be proved by the First Book thus: Erect CD at right angles to AB, and make CD = AC or CB. Join AD, DB. Then AD2 = AC2 + CD2 = 2AC2 In like manner,DB2 = 2CB2; therefore AD2 + DB2 = 2AC2 + 2CB2 = 4AC2. But since the angle ADB is right, AD2 + DB2 = AB2; thereforeAB2 = 4AC2. Cor. 3.—If a line be divided into any number of parts, the square on the whole is equal to the sum of the squares on all the parts, together with twice the sum of the rectangles contained by the several distinct pairs of parts. Exercises. 1. Prove Proposition iv. by using Propositions ii. and iii. 2. If from the vertical angle of a right-angled triangle a perpendicular be let fall on the hypotenuse, its square is equal to the rectangle contained by the segments of the hypotenuse. 3. From the hypotenuse of a right-angled triangle portions are cut off equal to the adjacent sides; prove that the square on the middle segment is equal to twice the rectangle contained by the extreme segments.