Circumcircle $ \mathcal P$ of $ \triangle DEF$ cuts line $ AD$ again at a point $ A'.$ Since $ DE = DF,$ line $ AD$ bisects both the angles $ \measuredangle EAF = \measuredangle CAB$ and $ \measuredangle EA'F.$ This is possible only when $ AD \perp EF,$ which means that $ \triangle AFE, \triangle ABC$ are A-isosceles, or when points $ A' \equiv A$ are identical, which means that quadrilateral $ AFDE$ is cyclic. The first possibility is eliminated by the condition that $ \triangle ABC$ is scalene. _____________________________________________________________________________________________ Circumcircle $ \mathcal P$ of $AFDE$ cuts $ BC$ at $ D$ and again at a point $ D'.$ Denote $ d = \overline{DD'}.$ On one hand, powers of $ B, C$ to $ \mathcal P$ are $ \text{pow} (B \to \mathcal P) = \overline{BA} \cdot \overline {BF} = c \cdot \frac {ca}{a + b},$ $\quad \text{pow} (C \to \mathcal P) = \overline{CA} \cdot \overline {CE} = b \cdot \frac {ab}{c + a}.$ On the other hand, $ \text{pow} (B \to \mathcal P) = \overline{BD} \cdot \overline {BD'} = \frac {ca}{b + c} \cdot \left(\frac {ca}{b + c} \pm d\right),$ $\quad \text{pow} (C \to \mathcal P) = \overline{CD} \cdot \overline {CD'} = \frac {ab}{b + c} \cdot \left(\frac {ab}{b + c} \mp d\right),$ where upper signs apply for $ b < c$ and lower signs for $ b > c$. _____________________________________________________________________________________________ Comparing the two expressions for $ \text{pow} (B \to \mathcal P)$ and the two expressions for $ \text{pow} (C \to \mathcal P)$ $\implies$ $ \frac{ca}{b + c} \cdot \left(\frac {ca}{b + c} \pm d\right) = c \cdot \frac {ca}{a + b},$ $\quad \frac {ab}{b + c} \cdot \left(\frac {ab}{b + c} \mp d\right) = b \cdot \frac {ab}{c + a}.$ Reducing these equations by factors .$ ca,\ ab,$ .resp. $\implies$ $ \frac{1}{b + c} \cdot \left( \frac {ca}{b + c} \pm d \right) = \frac {c}{a + b},$ $\quad \frac{1}{b + c} \cdot \left( \frac {ab}{b + c} \mp d \right) = \frac {b}{c + a}.$ Adding the two equations to eliminate segment $ d$ $\implies$ $ \frac{1}{b + c} \cdot \left( (c + b) \cdot \frac {a}{b + c} \right) = \frac {c}{a + b} + \frac {b}{c + a}.$ Reducing left side of the last equation by factor $ (b + c) = (c + b)$ $\implies$ $ \boxed{ \frac {a}{b + c} = \frac {b}{c + a} + \frac {c}{a + b}} \ .$ _____________________________________________________________________________________________

Multiplying this through $\implies$ (details)

$ a^2c + a^3 + abc + a^2b = b^2a + b^3 + abc + b^2c + c^2b + c^3 + abc + c^2a.$ Collecting terms with factors .$ a^2,\ b^2,\ c^2$ $\implies$ $ a^2(c + a + b) + abc = b^2(a + b + c) + abc + c^2(b + c + a) + abc,$ $ a^2 = b^2 + c^2 + \frac {abc}{a + b + c} > b^2 + c^2.$ By reverse Pythagorean theorem, angle $\boxed {\measuredangle CAB > 90^\circ}$ is obtuse. _____________________________________________________________________________________________

Attachments:

it can be done using cosines lemma in $\triangle CED$ and $\triangle BFD$ on $DE,DF$,respectively. and of course doing a lot of calculations. (the solution isn't mine,but i have a sketch of it's proof in this way)

See the proposed problem P5 from here.