Dem.—On AB describe the square ABDF [I. xlvi.], and through C draw CE parallel to AF [I. xxxi.]. Now, since AB is equal to AF, the rectangle contained by AB and AC is equal to the rectangle contained by AF and AC; but AE is the rectangle contained by AF and AC. Hence the rectangle contained by AB and AC is equal to AE. In like manner the rectangle contained by AB and CB is equal to the figure CD. Therefore the sum of the two rectangles AB.AC, AB.CB is equal to the square on AB.
| Or thus: | AB | = AC + CB, | |||||||||
| and | AB | = AB. | |||||||||
| Hence, multiplying, we get | AB2 | = AB.AC + AB.CB. |
This Proposition is the particular case of i. when the divided and undivided lines are equal, hence it does not require a separate Demonstration.
PROP. III.—Theorem.
If a line (AB) be divided into two segments (at C), the rectangle contained by the whole line and either segment (CB) is equal to the square on that segment together with the rectangle contained by the segments.
Dem.—On BC describe the square BCDE [I. xlvi.]. Through A draw AF parallel to CD: produce ED to meet AF in F. Now since CB is equal to CD, the rectangle contained by AC, CB is equal to the rectangle contained by AC, CD; but the rectangle contained by AC, CD is the figure AD. Hence the rectangle AC.CB is equal to the figure AD, and the square on CB is the figure CE. Hence the rectangle AC.CB, together with the square on CB, is equal to the figure AE.
Again, since CB is equal to BE, the rectangle AB.CB is equal to the rectangle AB.BE; but the rectangle AB.BE is equal to the figure AE. Hence the rectangle AB.CB is equal to the figure AE. And since things which are equal to the same are equal to one another, the rectangle AC.CB, together with the square on CB, is equal to the rectangle AB.CB.
| Or thus: | AB | = AC + CB, | |||||||||
| CB | = CB. | ||||||||||
| Hence | AB.CB | = AC.CB + CB2. |