PROPOSITION 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.
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.
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.
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.
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.
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]
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.
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.
And
FC is equal to
FB; therefore the rectangle
AE,
EC together with the square on
EF is equal to the square on
FB.
For the same reason, also, the rectangle
DE,
EB together with the square on
FE is equal to the square on
FB.
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.
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.
Therefore etc.