1. Cut a line externally in “extreme and mean ratio.”
2. The difference between the squares on the segments of a line divided in “extreme and mean ratio” is equal to their rectangle.
3. In a right-angled triangle, if the square on one side be equal to the rectangle contained by the hypotenuse and the other side, the hypotenuse is cut in “extreme and mean ratio” by the perpendicular on it from the right angle.
4. If AB be cut in “extreme and mean ratio” at C, prove that
| (1) | AB2 + BC2 = 3AC2. | ||
| (2) | (AB + BC)2 = 5AC2. |
5. The three lines joining the pairs of points G, B; F, D; A, K, in the construction of Proposition xi., are parallel.
6. If CH intersect BE in O, AO is perpendicular to CH.
7. If CH be produced, it meets BF at right angles.
8. ABC is a right-angled triangle having AB = 2AC: if AH be made equal to the difference between BC and AC, AB is divided in “extreme and mean ratio” at H.
PROP. XII.—Theorem.