Kitap 3
35
But the square on FE is equal to the squares on EG, GF, and the square on FC is equal to the squares on CG, GF; [I. 47] therefore the rectangle AE, EC together with the square on FE is equal to the square on FC.
35
And FC is equal to FB; therefore the rectangle AE, EC together with the square on EF is equal to the square on FB.
35
For the same reason, also, the rectangle DE, EB together with the square on FE is equal to the square on FB.
35
But the rectangle AE, EC together with the square on FE was also proved equal to the square on FB; therefore the rectangle AE, EC together with the square on FE is equal to the rectangle DE, EB together with the square on FE.
35
Let the square on FE be subtracted from each; therefore the rectangle contained by AE, EC which remains is equal to the rectangle contained by DE, EB.
35
Therefore etc.
PROPOSITION 36.
36
If a point be taken outside a circle and from it there fall on the circle two straight lines, and if one of them cut the circle and the other touch it, the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference will be equal to the square on the tangent.
36
For let a point D be taken outside the circle ABC, and from D let the two straight lines DCA, DB fall on the circle ABC; let DCA cut the circle ABC and let BD touch it; I say that the rectangle contained by AD, DC is equal to the square on DB.
