Therefore DA is the greatest, and DE greater than DF and DF greater than DC.
Again, because MK, KD are greater than MD, [I, 20.
and MK is equal to MG [I. Definition 15.
the remainder KD is greater than the remainder GD,
that is, GD is less than KD.
And because MLD is a triangle, and from the points
M, D, the extremities of its side MD the straight lines
MK, DK are drawn to the point K within the triangle,
therefore MK, KD are less than ML, LD; [I. 21.
and MK is, equal to ML ; [I. Definition 15.
therefore the remainder KD is less than the remainder LD.
In the same manner it may be shewn that LD is less
than HD.
Therefore DG is the least, and DK less than DL, and DL
less than DH.
Also, there can be drawn two equal straight lines from the point D to the circumference, one on each side of the least line.
For, at the point M, in the straight line MD, make the
angle DMB equal to the angle DMK, [I. 23.
and join DB.
Then, because MK is equal to MB,
and MD is common to the two triangles KMD, BMD ;
the two sides KM, MD are equal to the two sides BM, MD,
each to each ;
and the angle DMK is equal to the angle DMB ; [Constr.
therefore the base DK is equal to the base DB. [I. 4.
But, besides DB, no other straight line can be drawn
from D to the circumference, equal to DK.
For, if it be possible, let DN be equal to DK.
Then, because DN is equal to DK,
and DB is also equal to DK,
therefore DB is equal to DN; [Axiom 1.
that is, a line nearer to the least is equal to one which is more remote ; which is impossible by what has been already shewn.
Wherefore, if any point be taken &c. q.e.d.
