Kitap 4
3
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]
3
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.
3
But the angles DEG, DEF are also equal to two right angles; [I. 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.
3
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]
3
Therefore the triangle LMN is equiangular with the triangle DEF; and it has been circumscribed about the circle ABC.
3
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 and commonly. is in fact divisible, καὶ διαιρεῖται, literally is actually divided.
PROPOSITION 4.
4
In a given triangle to inscribe a circle.
4
Let ABC be the given triangle; thus it is required to inscribe a circle in the triangle ABC.
4
Let the angles ABC, ACB be bisected by the straight lines BD, CD [I. 9], and let these meet one another at the point D; from D let DE, DF, DG be drawn perpendicular to the straight lines AB, BC, CA.
