Exercises.
1. Prove Proposition iv. by using Propositions ii. and iii.
2. If from the vertical angle of a right-angled triangle a perpendicular be let fall on the hypotenuse, its square is equal to the rectangle contained by the segments of the hypotenuse.
3. From the hypotenuse of a right-angled triangle portions are cut off equal to the adjacent sides; prove that the square on the middle segment is equal to twice the rectangle contained by the extreme segments.
4. In any right-angled triangle the square on the sum of the hypotenuse and perpendicular, from the right angle on the hypotenuse, exceeds the square on the sum of the sides by the square on the perpendicular.
5. The square on the perimeter of a right-angled triangle is equal to twice the rectangle contained by the sum of the hypotenuse and one side, and the sum of the hypotenuse and the other side.
PROP. V.—Theorem.
If a line (AB) be divided into two equal parts (at C), and also into two unequal parts (at D), the rectangle (AD.DB) contained by the unequal parts, together with the square on the part (CD) between the points of section, is equal to the square on half the line.
Dem.—On CB describe the square CBEF [I. xlvi.]. Join BF. Through D draw DG parallel to CF, meeting BF in H. Through H draw KM parallel to AB, and through A draw AK parallel to CL [I. xxxi.].