PROPOSITION 8.
If a point be taken outside a circle and from the point straight lines be drawn through to the circle,
one of which is through the centre and the others are drawn at random,
then,
of the straight lines which fall on the concave circumference,
that through the centre is greatest,
while of the rest
the nearer to that through the centre is always greater than the more remote,
but,
of the straight lines falling on the convex circumference,
that between the point and the diameter is least,
while of the rest the nearer to the least is always less than the more remote,
and only two equal straight lines will fall on the circle from the point,
one on each side of the least.
Let
ABC be a circle, and let a point
D be taken outside
ABC; let there be drawn through from it straight lines
DA,
DE,
DF,
DC, and let
DA be through the centre; I say that, of the straight lines falling on the concave circumference
AEFC, the straight line
DA through the centre is greatest, while
DE is greater than
DF and
DF than
DC.; but, of the straight lines falling on the convex circumference
HLKG, the straight line
DG between the point and the diameter
AG is least; and the nearer to the least
DG is always less than the more remote, namely
DK than
DL, and
DL than
DH.
For let the centre of the circle
ABC be taken [
III. 1], and let it be
M; let
ME,
MF,
MC,
MK,
ML,
MH be joined.
Then, since
AM is equal to
EM, let
MD be added to each;
therefore AD is equal to EM, MD.
But
EM,
MD are greater than
ED; [
I. 20]
therefore AD is also greater than ED.
Again, since
ME is equal to
MF,
and MD is common, therefore
EM,
MD are equal to
FM,
MD;
and the angle EMD is greater than the angle FMD;
therefore the base ED is greater than the base FD. [I. 24]
Similarly we can prove that
FD is greater than
CD; therefore
DA is greatest, while
DE is greater than
DF, and
DF than
DC.
Next, since
MK,
KD are greater than
MD, [
I. 20] and
MG is equal to
MK, therefore the remainder
KD is greater than the remainder
GD,
so that GD is less than KD.
And, since on
MD, one of the sides of the triangle
MLD, two straight lines
MK,
KD were constructed meeting within the triangle, therefore
MK,
KD are less than
ML,
LD; [
I. 21] and
MK is equal to
ML;
therefore the remainder DK is less than the remainder DL.
Similarly we can prove that
DL is also less than
DH;
therefore DG is least, while DK is less than DL, and DL than DH.
I say also that only two equal straight lines will fall from the point
D on the circle, one on each side of the least
DG.
On the straight line
MD, and at the point
M on it, let the angle
DMB be constructed equal to the angle
KMD, and let
DB be joined.
Then, since
MK is equal to
MB, and
MD is common,
the two sides KM, MD are equal to the two sides BM, MD respectively; and the angle
KMD is equal to the angle
BMD;
therefore the base DK is equal to the base DB. [I. 4]
I say that no other straight line equal to the straight line
DK will fall on the circle from the point
D.
For, if possible, let a straight line so fall, and let it be
DN.
Then, since DK is equal to DN,
while
DK is equal to
DB,
DB is also equal to DN, that is, the nearer to the least
DG equal to the more remote: which was proved impossible.
Therefore no more than two equal straight lines will fall on the circle
ABC from the point
D, one on each side of
DG the least.
Therefore etc.
