Or thus: Produce BA to D, and make AD = BC. On DB describe the square DBEF. Cut off BG, EI, FL each equal to BC. Through A and I draw lines parallel to DF, and through G and L, lines parallel to AB.
Now it is evident that the four rectangles. AG, GI, IL, LA are all equal; but AG is the rectangle AB.BG or AB.BC. Therefore the sum of the four rectangles is equal to 4AB.BC. Again, the figure NP is evidently equal to the square on AC. Hence the whole figure, which is the square on BD, or the square on the sum of AB and BC, is equal to 4AB.BC + AC2.
| Or thus: | AB + BC | = AC + 2BC; | |||||||||
| therefore | (AB + BC)2 | = AC2 + 4AC.CB + 4BC2 | |||||||||
| = AC2 + 4(AC + CB).CB | |||||||||||
| = AC2 + 4AB.BC. |
Direct sequence from v. or vi.
Since by v. or vi. the rectangle contained by any two lines is = the square on half their sum − the square on half their difference; therefore four times the rectangle contained by any two lines = the square on their sum − the square on their difference.
Direct sequence of viii. from iv. and vii.
By iv., the square on the sum = the sum of the squares + twice the rectangle.
By vii., the square on the difference = the sum of the squares − twice the rectangle. Therefore, by subtraction, the square on the sum − the square on the difference = four times the rectangle.
Exercises.
1. In the figure [I. xlvii.] if EF, GK be joined, prove EF2 − CO2 = (AB + BO)2.