I think the problem statement is wrong - the points $D$ and $E$ should be defined as $CI\cap AB$ and $BI\cap AC$ respectively. Anyway, if my formulation is right, here is the solution. (a) We use directed angles modulo $180^{\circ}$. $$\measuredangle(DK,BC)=\measuredangle(DK,DC)+\measuredangle(DC,BC)=\measuredangle(DC,DA)+\measuredangle(AC,DC)=\measuredangle(AC,DA)=\measuredangle CAB$$Similarly, $\measuredangle(KE,BC)=\measuredangle BAC$. It thus suffices to show that $IK$ bisects $\angle DKE$. $$\frac{\sin\angle DKI}{\sin\angle IKE}=\frac{DI}{IE}\cdot\frac{\sin\angle KDI}{\sin\angle KEI}=\frac{DI}{IE}\cdot\frac{\sin\angle BDI}{\sin\angle CEI}=\frac{DI}{IE}\cdot\frac{\frac{\sin\angle DBI}{DI}\cdot BI}{\frac{\sin\angle ECI}{EI}\cdot CI}=\frac{\sin\angle IBC}{\sin\angle ICB}\cdot\frac{BI}{CI}=1.$$Clearly we have $\angle DKI\neq 180^{\circ}-\angle IKE$. Thus $\angle DKI=\angle IKE$ and we're done. (b) If $IDKE$ is cyclic, then $\angle DKE=180^{\circ}-2\angle A$, and also equal to $180^{\circ}-\angle DIE=90^{\circ}-\frac{1}{2}\angle A$. Hence, $2\angle A=90^{\circ}+\frac{1}{2}\angle A$, which means that $\angle A=60^{\circ}$. Now $$BD+CE=\frac{BC}{AC+BC}\cdot AB+\frac{BC}{AB+BC}\cdot AC=BC\left(\frac{c}{a+b}+\frac{b}{a+c}\right).$$Now it suffices to show that $\frac{c}{a+b}+\frac{b}{a+c}=1$, or equivalently, $c(a+c)+b(a+b)=(a+b)(a+c)$. This is equivalent to showing that $b^2+c^2=a^2+bc$, which is true by the cosine rule.

Quote: Let $ABC$ be a triangle whose incircle $(I)$ is tangent to $AB, AC$ at $D, E$ respectively. Denote by $\Delta_b,\Delta_c$ the lines symmetric to the lines $AB, AC$ with respect to $CD, BE$ correspondingly. Suppose that $\Delta_b,\Delta_c$ meet at $K$. a) Prove that $IK \perp BC$. b) If $I \in (K DE)$, prove that $BD + C E = BC$. that above was the official wrong statement, thanks for noticing the official solution, proves you correct, I just edited it to Quote: Let $ABC$ be a triangle with incenter $I$ . Let $CI, BI$ intersect $AB, AC$ at $D, E$ respectively. Denote by $\Delta_b,\Delta_c$ the lines symmetric to the lines $AB, AC$ with respect to $CD, BE$ correspondingly. Suppose that $\Delta_b,\Delta_c$ meet at $K$. a) Prove that $IK \perp BC$. b) If $I \in (K DE)$, prove that $BD + C E = BC$.