PROPOSITION 37.
If a point be taken outside a circle and from the point there fall on the circle two straight lines,
if one of them cut the circle,
and the other fall on it,
and if further 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 be equal to the square on the straight line which falls on the circle,
the straight line which falls on it will touch the circle.
For let a point
D be taken outside the circle
ABC; from
D let the two straight lines
DCA,
DB fall on the circle
ACB; let
DCA cut the circle and
DB fall on it; and let the rectangle
AD,
DC be equal to the square on
DB.
I say that
DB touches the circle
ABC.
For let
DE be drawn touching
ABC; let the centre of the circle
ABC be taken, and let it be
F; let
FE,
FB,
FD be joined.
Thus the angle
FED is right. [
III. 18]
Now, since
DE touches the circle
ABC, and
DCA cuts it, the rectangle
AD,
DC is equal to the square on
DE. [
III. 36]
But the rectangle
AD,
DC was also equal to the square on
DB; therefore the square on
DE is equal to the square on
DB;
therefore DE is equal to DB.
And
FE is equal to
FB; therefore the two sides
DE,
EF are equal to the two sides
DB,
BF; and
FD is the common base of the triangles;
therefore the angle DEF is equal to the angle DBF. [I. 8]
But the angle
DEF is right;
therefore the angle DBF is also right.
And
FB produced is a diameter; and the straight line drawn at right angles to the diameter of a circle, from its extremity, touches the circle; [
III. 16, Por.]
therefore DB touches the circle.
Similarly this can be proved to be the case even if the centre be on
AC.
Therefore etc. Q. E. D.
