As I don't know any synthetic geometry trick to solve this problem, my approach to tackle it is an analytical one.
I'm going to calculate the equation of line DE and the equations of circumcircle and incircle of triangle ABC. From the equations of these circles I will get the equation of the radical axis, and then it will be easy to show that OI is perpendicular to line DE
Let the point C be the origin of cartesian oblique axes, so that line ACE becomes the x-axis and line BCD becomes the y-axis. Then the coordinates of C are $C=(0,0)$ and the coordinates of B and A can be, without loss of generality, $B=(0,q)$ and $A=(p,0)$, with $p<0$ and $q<0$. Then, if we call c the measure of side AB, the coordinates of D and E are $D=(0,q+c)$ and $E=(p+c,0)$.
With these coordinates, we immediately note that line DE has the equation
The equation of the circumcircle, which passes through points A, B, C, is obviously
$$x^2+2xy\cos C +y^2 -px -qy=0$$
As for the incircle, it's not hard to see that it touches the x-axis and y-axis in the points $((p+q+c)/2,0)$ and $(0,(p+q+c)/2)$, which leads us to almost effortlessly arrive at its equation:
$$x^2+2xy\cos C +y^2 -(p+q+c)x -(p+q+c)y +(1/4)(p+q+c)^2=0$$
Subtracting both equations of these circles we finally get the equation of their radical axis:
Comparing it with the equation of DE we have previously got, we see that both are parallel to each other, so that OI is perpendicular to DE.
This problem, by the way, shows us two important points:
i) we shouldn't disparage or forget the cartesian oblique axes;
ii) how a wise choice of a coordinate system, axes and origin can pave the way to a smooth solution.