8. If through a point O within a triangle ABC parallels EF, GH, IK to the sides be drawn, the sum of the rectangles of their segments is equal to the rectangle contained by the segments of any chord of the circumscribing circle passing through O.
| Dem.—AO.AL = AB.AK + AC.AE. | (6) | ||||||||||
| But | AO2 = AG.AK + AH.AE − GO.OH. | (7) | |||||||||
| Hence | AO.OL = BG.AK + CH.AE + GO.OH, | ||||||||||
| or | AO.OL = EO.OF + IO.OK + GO.OH. |
9. The rectangle contained by the side of an inscribed square standing on the base of a triangle, and the sum of the base and altitude, is equal to twice the area of the triangle.
10. The rectangle contained by the side of an escribed square standing on the base of a triangle, and the difference between the base and altitude, is equal to twice the area of the triangle.
11. If from any point P in the circumference of a circle a perpendicular be drawn to any chord, its square is equal to the rectangle contained by the perpendiculars from the extremities of the chord on the tangent at P.
12. If O be the point of intersection of the diagonals of a cyclic quadrilateral ABCD, the four rectangles AB.BC, BD.CD, CD.DA, DA.AB, are proportional to the four lines BO, CO, DO, AO.
13. The sum of the rectangles of the opposite sides of a cyclic quadrilateral ABCD is equal to the rectangle contained by its diagonals.
Dem.—Make the angle DAO = CAB; then the triangles DAO, CAB are equiangular; therefore AD : DO :: AC : CB; therefore AD.BC = AC.DO. Again, the triangles DAC, OAB are equiangular, and CD : AC :: BO : AB; therefore AC.CD = AC.BO. Hence AD.BC+AB.CD = AC.BD.3[ 3]This Proposition is known as Ptolemy’s theorem.