The First Six Books of the Elements of Euclid - Euclid; John Casey

Prop. iii. is the particular case of Prop. i., when the undivided line is equal to a segment of the divided line.

PROP. IV.—Theorem.

If a line (AB) be divided into any two parts (at C), the square on the whole line is equal to the sum of the squares on the parts (AC, CB), together with twice their rectangle.

Dem.—On AB describe a square ABDE. Join EB; through C draw CF parallel to AE, intersecting BE in G; and through G draw HI parallel to AB.

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.