Dem.—From the vertex C let fall the perpendicular CD. Draw EF parallel to AB, and AE, BF each parallel to CD. Then AF is the rectangle contained by AB and BF; but BF is equal to CD. Hence AF = AB.CD; but [I. xli.] the triangle ABC is = half the parallelogram AF. Therefore the triangle ABC is = AB.CD. PROP. II.—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 rectangles contained by the whole and each of the segments (AC, CB). 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. 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.