Given a segment of a circle
, to describe the complete circle of which it is a segment
be the given segment of a circle; thus it is required to describe the complete circle belonging to the segment ABC
, that is, of which it is a segment.
For let AC
be bisected at D
, let DB
be drawn from the point D
at right angles to AC
, and let AB
. be joined; the angle ABD is then greater than, equal to, or less than the angle BAD.
First let it be greater; and on the straight line BA
, and at the point A
on it, let the angle BAE
be constructed equal to the angle ABD
; let DB
be drawn through to E
, and let EC
Then, since the angle ABE
is equal to the angle BAE
the straight line EB is also equal to EA. [I. 6]
And, since AD
is equal to DC
, and DE
is common, the two sides AD, DE are equal to the two sides CD, DE respectively;
and the angle ADE
is equal to the angle CDE
, for each is right; therefore the base AE is equal to the base CE.
was proved equal to BE
; therefore BE is also equal to CE;
therefore the three straight lines AE
are equal to one another.
Therefore the circle drawn with centre E
and distance one of the straight lines AE
will also pass through the remaining points and will have been completed. [III. 9
Therefore, given a segment of a circle, the complete circle has been described.
And it is manifest that the segment ABC
is less than a semicircle, because the centre E
happens to be outside it.
Similarly, even if the angle ABD
be equal to the angle BAD
being equal to each of the two BD
, the three straight lines DA, DB, DC will be equal to one another,
D will be the centre of the completed circle, and ABC will clearly be a semicircle.
But, if the angle ABD
be less than the angle BAD
, and if we construct, on the straight line BA
and at the point A
on it, an angle equal to the angle ABD
, the centre will fall on DB
within the segment ABC
, and the segment
will clearly be greater than a semicircle.
Therefore, given a segment of a circle, the complete circle has been described. Q. E. F.