Proposition 4.4 - In a given triangle, inscribe a circle.

Proof
Let ABC be a triangle. Let <ABC and <ACB be bisected and the point of interesection of the angle bisectors be named D. Now, draw three lines DE, DF, and DG such that these lines are perpendicular to AB, BC, and AC respectively. Using D as the center and DF as the radius, draw a circle.
Consider triangles EBD and DBF. <ABD=<DBF because DB is the angle bisector, <BED=<DFB because they are both right, and DB=BD. By AAS, triangles EBD=DBF. Thus, ED=DF. Similarily, triangles DFC=GDC and DF=DG. Now we have that DF=DE=DG. Since DF is the radius of the drawn circle, DE and DG are also radii.
Now, circle EFG touches triangle ABC at E,F, and G only by propositions 3.16 and 4.5. Therefore, EFG is inscribed in ABC.