Let $AD$, $BE$ and $CF$ be the angle bisectors of triangle $ABC$. Let the points $D$, $E$ and $F$ lie on $BC$, $AC$ and $AB$, respectively. Let $I$ denote the incenter of $ABC$.
Prove that if the sum of the areas of the inside triangles $ICE$, $IAF$ and $IBF$ equals half the area of triangle $ABC$, then $ABC$ is isosceles. In other words:
$$\text{area}(ICE) + \text{area}(IAF) + \text{area}(IBF) = \frac{1}{2} \text{area} (ABC)\Rightarrow ABC~ \text{is isosceles}.$$
Using $S = a b \sin(\gamma)/2$ and the angle bisector theorem, we have $$\frac {S_1} {S_2} = \frac {S_{IAB}} {S_{IBC}} = \frac {AB} {BC} = k, \\ \frac {S_4} {S_3} = \frac {S_{IAE}} {S_{ICE}} = \frac {AE} {CE} = k, \\ (S_1 + S_3) - (S_2 + S_4) = (k S_2 + S_3) - (S_2 + k S_3) = \\ {(k - 1) (S_2 - S_3)} = 0.$$ The first possibility is $k = 1$, giving the required result (then the difference $S_2 - S_3$ is arbitrary). The second possibility is $$S_2 = S_3 \Rightarrow BC = CE, \\ S_1 = S_4 \Rightarrow AB = AE, \\ AB + BC = AE + CE = AC,$$ giving a degenerate triangle.