Kitap 3
34
Therefore from the given circle ABC the segment BAC. has been cut off admitting an angle equal to the given rectilineal angle, the angle at D. Q. E. F.
PROPOSITION 35.
35
If in a circle two straight lines cut one another, the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other.
35
For in the circle ABCD let the two straight lines AC, BD cut one another at the point E; I say that the rectangle contained by AE, EC is equal to the rectangle contained by DE, EB.
35
If now AC, BD are through the centre, so that E is the centre of the circle ABCD, it is manifest that, AE, EC, DE, EB being equal, the rectangle contained by AE, EC is also equal to the rectangle contained by DE, EB.
35
Next let AC, DB not be through the centre; let the centre of ABCD be taken, and let it be F; from F let FG, FH be drawn perpendicular to the straight lines AC, DB, and let FB, FC, FE be joined.
35
Then, since a straight line GF through the centre cuts a straight line AC not through the centre at right angles, it also bisects it; [III. 3] therefore AG is equal to GC.
35
Since, then, the straight line AC has been cut into equal parts at G and into unequal parts at E, the rectangle contained by AE, EC together with the square on EG is equal to the square on GC; [II. 5]
35
Let the square on GF be added; therefore the rectangle AE, EC together with the squares on GE, GF is equal to the squares on CG, GF.
