Dem.—On AB describe the square ABDE. Join BE. Through C draw CG parallel to AE, intersecting BE in F. Through F draw HK parallel to AB.
Now the square AD is equal to the three figures AK, FD, and GH: to each add the square CK, and we have the sum of the squares AD, CK equal to the sum of the three figures AK, CD, GH; but CD is equal to AK; therefore the sum of the squares AD, CK is equal to twice the figure AK, together with the figure GH. Now AK is the rectangle AB.BK; but BK is equal to BC; therefore AK is equal to the rectangle AB.BC, and AD is the square on AB; CK the square on CB; and GH is the square on HF, and therefore equal to the square on AC. Hence the sum of the squares on AB and BC is equal to twice the rectangle AB.BC, together with the square on AC.
Or thus: On AC describe the square ACDE. Produce the sides CD, DE, EA, and make each produced part equal to CB. Join BF, FG, GH, HB. Then the figure BFGH is a square [I. xlvi., Ex. 3], and it is equal to the square on AC, together with the four equal triangles HAB, BCF, FDG, GEH. Now [I. xlvii.], the figure BFGH is equal to the sum of the squares on AB, AH—that is, equal to the sum of the squares on AB, BC; and the sum of the four triangles is equal to twice the rectangle AB.BC, for each triangle is equal to half the rectangle AB.BC. Hence the sum of the squares on AB, BC is equal to twice the rectangle AB.BC, together with the square on AC.
| Or thus: | AC = AB − BC; | ||||||||||
| therefore | AC2 = AB2 − 2AB.BC | + BC2; | |||||||||
| therefore | AC2 + 2AB.BC = AB2 | + BC2. |
Comparison of iv. and vii.
By iv., square on sum = sum of squares + twice rectangle.
By vii., square on difference = sum of squares-twice rectangle.
Cors. from iv. and vii.