I found it !
Okay look at the sketch. It is not necessary to draw the inscribed circles.
Consider triangles DEM and ABC. All their sides are parallel to each other. Hence they're congruent (or similar). Let be the ratio between the two. This means that
This always holds true for two congruent triangles.
It'll be the same for triangle FGM :
and for triangle HMI :
Now consider the blue parts of the equalities.
You can see that if you sum them, it should give 1, according to the result you're given.
So we'll prove that the sum of the red parts is equal to 1 !
But since EM//BH and EB//MH (parallel), then EMHB is a parallelogram. Hence
Similarly, MGCI is a parallelogram. And hence
So the red sum becomes :
you can see that the points B, H, I and C are aligned. Hence
So the red part is equal to 1.
Therefore for the blue part :
And you're done