Let EF be produced in both directions to the points G , H , let the centre K of the circle ABC be taken [[III. 1](elem.3.1)], and let the straight line KB be drawn across at random; on the straight line KB , and at the point K on it, let the angle BKA be constructed equal to the angle DEG , and the angle BKC equal to the angle DFH ; [[I. 23](elem.1.23)] and through the points A , B , C let LAM , MBN , NCL be drawn touching the circle ABC . [[III. 16, Por.](elem.3.16.p.1)]

Now, since LM , MN , NL touch the circle ABC at the points A , B , C , and KA , KB , KC have been joined from the centre K to the points A , B , C , therefore the angles at the points A , B , C are right. [[III. 18](elem.3.18)]

And, since the four angles of the quadrilateral AMBK are equal to four right angles, inasmuch as AMBK is in fact divisible into two triangles, and the angles KAM , KBM are right,
therefore the remaining angles AKB , AMB are equal to two right angles.

But the angles DEG , DEF are also equal to two right angles; [[I. 13](elem.1.13)] therefore the angles AKB , AMB are equal to the angles DEG , DEF , of which the angle AKB is equal to the angle DEG ; therefore the angle AMB which remains is equal to the angle DEF which remains.

Similarly it can be proved that the angle LNB is also equal to the angle DFE ; therefore the remaining angle MLN is equal to the angle EDF . [[I. 32](elem.1.32)]

Therefore the triangle LMN is equiangular with the triangle DEF ; and it has been circumscribed about the circle ABC .

Therefore about a given circle there has been circumscribed a triangle equiangular with the given triangle. Q. E. F.

at random, literally as it may chance,

ὡς ἕτυχεν . The same expression is used in [III. 1](elem.3.1) and commonly.

is in fact divisible, καὶ διαιρεῖται , literally is actually divided.