PROPOSITION 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.
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.
Then
DCA is either through the centre or not through the centre.
First let it be through the centre, and let
F be the centre of the circle
ABC; let
FB be joined;
therefore the angle FBD is right. [III. 18]
And, since
AC has been bisected at
F, and
CD is added to it, the rectangle
AD,
DC together with the square on
FC is equal to the square on
FD. [
II. 6]
But
FC is equal to
FB; therefore the rectangle
AD,
DC together with the square on
FB is equal to the square on
FD.
And the squares on
FB,
BD are equal to the square on
FD; [
I. 47] therefore the rectangle
AD,
DC together with the square on
FB is equal to the squares on
FB,
BD.
Let the square on
FB be subtracted from each; therefore the rectangle
AD,
DC which remains is equal to the square on the tangent
DB.
Again, let
DCA not be through the centre of the circle
ABC; let the centre
E be taken, and from
E let
EF be drawn perpendicular to
AC; let
EB,
EC,
ED be joined.
Then the angle
EBD is right. [
III. 18]
And, since a straight line
EF through the centre cuts a straight line
AC not through the centre at right angles,
it also bisects it; [III. 3] therefore AF is equal to FC.
Now, since the straight line
AC has been bisected at the point
F, and
CD is added to it, the rectangle contained by
AD,
DC together with the square on
FC is equal to the square on
FD. [
II. 6]
Let the square on
FE be added to each; therefore the rectangle
AD,
DC together with the squares on
CF,
FE is equal to the squares on
FD,
FE.
But the square on
EC is equal to the squares on
CF,
FE, for the angle
EFC is right; [
I. 47] and the square on
ED is equal to the squares on
DF,
FE; therefore the rectangle
AD,
DC together with the square on
EC is equal to the square on
ED.
And
EC is equal to
EB; therefore the rectangle
AD,
DC together with the square on
EB is equal to the square on
ED.
But the squares on
EB,
BD are equal to the square on
ED, for the angle
EBD is right; [
I. 47] therefore the rectangle
AD,
DC together with the square on
EB is equal to the squares on
EB,
BD.
Let the square on
EB be subtracted from each; therefore the rectangle
AD,
DC which remains is equal to the square on
DB.
Therefore etc. Q. E. D.