PROPOSITION 14.
In a circle equal straight lines are equally distant from the centre,
and those which are equally distant from the centre are equal to one another.
Let
ABDC be a circle, and let
AB,
CD be equal straight lines in it; I say that
AB,
CD are equally distant from the centre.
For let the centre of the circle
ABDC be taken [
III. 1], and let it be
E; from
E let
EF,
EG be drawn perpendicular to
AB,
CD, and let
AE,
EC be joined.
Then, since a straight line
EF through the centre cuts a straight line
AB not through the centre at right angles, it also bisects it. [
III. 3]
Therefore AF is equal to FB; therefore AB is double of AF.
For the same reason
CD is also double of CG; and
AB is equal to
CD;
therefore AF is also equal to CG.
And, since
AE is equal to
EC,
the square on AE is also equal to the square on EC. But the squares on
AF,
EF are equal to the square on
AE, for the angle at
F is right; and the squares on
EG,
GC are equal to the square on
EC, for the angle at
G is right; [
I. 47]
therefore the squares on AF, FE are equal to the squares on CG, GE, of which the square on
AF is equal to the square on
CG, for
AF is equal to
CG;
therefore the square on FE which remains is equal to the square on EG, therefore EF is equal to EG.
But in a circle straight lines are said to be equally distant from the centre when the perpendiculars drawn to them from the centre are equal; [
III. Def. 4]
therefore AB, CD are equally distant from the centre.
Next, let the straight lines
AB,
CD be equally distant from the centre; that is, let
EF be equal to
EG.
I say that
AB is also equal to
CD.
For, with the same construction, we can prove, similarly, that
AB is double of
AF, and
CD of
CG.
And, since
AE is equal to
CE,
the square on AE is equal to the square on CE. But the squares on
EF,
FA are equal to the square on
AE, and the squares on
EG,
GC equal to the square on
CE. [
I. 47]
Therefore the squares on
EF,
FA are equal to the squares on
EG,
GC, of which the square on
EF is equal to the square on
EG, for
EF is equal to
EG; therefore the square on
AF which remains is equal to the square on
CG;
therefore AF is equal to CG. And
AB is double of
AF, and
CD double of
CG;
therefore AB is equal to CD.
Therefore etc. Q. E. D.
