Let d be the common length |AB| = |CD| = |EF| = d.

Let "O" be the center of the given circle.
Connect this point O with the intersection points C and D by the radii OC and OD.
The triangle COD is isosceles (its sides OC and OD are the radii !). 
Draw the perpendicular OM from O to CD. Then OM is the altitude in the isosceles triangle COD; hence, OM is the median, too.
It means that CM = DM = .

Calculate the length of OM from the right-angled triangle MOD:

|OM| = .

If you make similar construction/drawing and calculations for two other triangles OAB and OEF, you will see that

their altitudes are expressed by the same formula, and, hence, have THE SAME length.

Thus the point O is equidistant from the three sides of the given triangle.

Hence, the point O is the center of the inscribed circle to the given triangle PQR.