Anybody?

We begin with a preliminary result. We will work with unoriented angles, i.e. the solution is configuration-dependent. Lemma 1. The magnitude of angle $ \angle{KID}$ is $ \frac {B - C}{2}$. Proof of Lemma 1. We will give a metric proof different from the official ones. Denote by $ A^{\prime}$ the antipode of $ A$ with respect to the circumcircle $ (O)$ of triangle $ ABC$. In this case, triangle $ AKA^{\prime}$ is isosceles and since the cevians $ AH$ and $ AO$ are isogonal with respect to the angle $ \angle{BAC}$ we deduce that $ I$ lies on the line bisector $ AM$ of the segment $ A^{\prime}K$ and moreover, the midpoint $ M$ of $ A^{\prime}K$ is the midpoint of the small arc $ BC$. Since $ \angle{DAI} = \angle{DAO}/2 = (B - C)/2$, it is suffice to prove that $ \angle{KID} = \angle{DAI}$, which is equivalent with the fact that the line $ KI$ is tangenct to the circumcircle of triangle $ DAI$ at $ I$. In this case, we will resume to proving that $ KA \cdot KD = KI^{2}$, which will yield our conclusion. As I was saying, $ KI^{2} = AK^{2} + AI^{2} - 2AI \cdot AK \cdot \cos{\frac {B - C}{2}}$ $ = 4R^{2} + \frac {r^{2}}{\sin^{2}{\frac {A}{2}}} - \frac {4Rr\cos{\frac {B - C}{2}}}{\sin{\frac {A}{2}}}$ $ = 4R^{2} + \frac {r^{2}bc}{(p - b)(p - c)} - \frac {4Rr(b + c)}{a}$ $ = 4R^{2} + \frac {bc(p - a)}{p} - \frac {4Rr(b + c)}{a}$ $ = 4R^{2} - \frac {4pRr(b + c) - abc(p - a)}{ap}$ $ = 4R^{2} - \frac {4RS}{a}$ $ = 2R(2R - h_{a}) = KA \cdot KD$. This proves Lemma 1. We will end this proof apparently in the same manner as the official proof. Denote by $ A_{1}$, $ B_{1}$ the tangency points of the incircle with the sides $ BC$ and $ CA$, respectively. Since the two triangles $ IEB_{1}$ and $ IFA_{1}$ are congruent, $ \angle{IEB_{1}} = \angle{IFA_{1}}$. Now since $ \angle{B} \geq \angle{C}$, $ A_{1} \in (CD)$ and $ F \in A_{1}D$. Thus, $ \angle{IFC}$ is acute. We distinguish two cases, depending on the position of $ B_{1}$ and $ E$ on the side $ AC$. (1) $ A - B_{1} - E - C$: We have that $ \angle{IFC} = \angle{IEA}$, and thus the quadrilateral $ CEIF$ is cyclic which implies that $ \angle{FCE}( = \angle{C}) = \angle{KID} = (B - C)/2$ and so $ B = 3C$. (2) $ A - E - B_{1} - C$: In this case $ CEIF$ is a deltoid such that $ \angle{IEC} = \angle{IFC} < \pi/2$. Then we have $ \angle{FCE}( = \angle{C}) > 180^{\circ} - \angle{EIF} = \angle{KID} = (B - C)/2$. Hence, $ \angle{B} < 3\angle{C}$. This proves our problem.

Two more proofs of pohoatza's lemma:

let$ I_{a}$ is $A-$excenter. let $AI$ meet $BC$ at $S$. $AI.AI_{a}=AB.AC=2R.AD=AD.AK$ $\widehat{KII_{a}}=\widehat{I_{a}DK}.(A,S,I,I_{a})=-1\Rightarrow \widehat{KDS}=\widehat{IDS}=90-\widehat{FII_{a}}$ case 1: $\widehat{IFC}=\widehat{IEC}$ easy by angle chasing to find $B=3C$ case 2: $\widehat{IFC}+\widehat{IEC}=180$ easy by Calculation angle $\widehat{DKI}$

I give a proof of the fact that $\angle KID=\frac{\angle B-\angle C}{2}$ and the rest is the same as the proof given by Pohoatza. Let $A'$ be the antipode of $A,$ $J$ be the $A$-excenter for triangle $ABC$ and $M$ the midpoint of $IJ$. Note that $\angle AMA'=90^{\circ}$ and from the fact that $\triangle AIC \sim \triangle ABJ$ we have $AI\cdot AJ=AB\cdot AC$. Also, $AD\cdot AK=AD\cdot AA'=AB\cdot AC$ where the last equality follows since $\triangle ABD \sim \triangle AA'C$. Finally, as \begin{align*} KD\cdot KA=AK^2-AK\cdot AD=AA'^2-AI\cdot AJ &= AA'^2-(AM^2-MI^2) \\ &=(AA'^2-AM^2)+MI^2\\ &=A'M^2+MI^2=A'I^2=KI^2 \end{align*}we get $\triangle KID \sim \triangle KAI \Longrightarrow \angle KID=\angle DAI$ and the result follows. Note The most difficult part of this problem was to actually "guess" the result we showed here since it is not a very "well-known" statement.

Notice that $AD$ lies on the same side of the angle bisector as $B$.So upon assuming that $\angle IFC$ is obtuse we obtain that $\measuredangle(AI,BC)$ is acute a contradiction and hence $\angle AFC$ is acute. Let $I_a$ be the excenter ,$A'$ the antipode.Notice tha $\sqrt{bc}$ inversion $AA'\cdot AD=bc=AD\cdot AK=AI\cdot AI_1$ $\implies$ $IDKI_A$ is cyclic and as $KI=KI_A$ upon inverting thru $K$ with radius $KI$ we obtain $KI^2=KD\cdot KA$ $\implies$ $KI$ is tangent to $\odot AID$ $\implies$ $\angle DIK=\angle DAI=\frac{\beta-\gamma}{2}$. If one of $\angle CEI,\angle CFI$ is obtuse ,by first paragraph, other one is acute $\implies$ in this case $AIFE $is cyclic and so $\gamma=\frac{\beta-\gamma}{2}$.So assume they're both acute (and hence equal): $$\angle CDI=\angle CFI-\angle FID=\pi-\gamma-\angle CFI \implies \gamma=\pi-2\angle CFI+\angle FID$$$\gamma\geq \frac{\beta-\gamma}{2}$ and hence the desired.$\blacksquare$

Let $A'$ be the $A$-antipode. Clearly, since $AK=AA'$, and $AI$ is the angle bisector of $\angle DAO$, when $AI$ intersects $KA'$ at $L$, $AL$ is also the altitude. Thus, $L$ lies on $(ABC)$. In fact, it is the arc midpoint of $BC$. Thus, by incenter-excenter lemma, $L$ is the center of a circle $(BIC)$. Since $\triangle ABD\sim \triangle AA'C$, we have $AD\cdot AA'=AB\cdot AC$. Let $B'\neq C$ be the point of intersect of circle $L$ with $AC$. Clearly, $AB=AB'$. We have \[KI^2 = A'I^2 = A'A^2 - AL^2 + IL^2 = A'A^2 - \text{Pow}_{(BIC)}(A) = A'A^2 - AB\cdot AC\]On the other hand, \[KA\cdot KD=KA^2 - AK\cdot AD=A'A^2-AA'\cdot AD=A'A^2-AB\cdot AC\]so $KI^2=KA\cdot KD$. We have $\angle KID=\angle KAI=\tfrac{\angle B - \angle C}{2}$. Let $H_A$ and $H_B$ be the feet of the altitudes from $I$ to $BC$ and $AC$, respectively. Note that $F$ must be on the segment $BH_A$. Let $E'$, regardless of the position of $E$, be the point on segment $CH_B$ for which $IE'=IF$. Evidently, $IF=IE'$ implies $\angle H_BIE'=\angle H_AIF$ so $\angle H_BIH_A=\angle FIE'=180^\circ-\angle C$. Thus, \[\angle DIF=180^\circ-\angle FIE\ge 180^\circ-\angle FIE'=\angle C\]and we are done.

Clean! Let $A'$ be the antipode of $A$ and $M$ the arc midpoint; we take two cases. Case 1: $CIEF$ is cyclic. Here, we can directly angle chase: note that \[\angle AIA'=\angle AIK = \angle AIB+\angle BID + \angle DIK = 90 + 3\angle C/2 + \angle BID\]Now, I claim that $\angle BID = \angle CIA'$. This is since, letting $S$ be sharkydevil and $P$ $AD\cap (AI)$, we have \[\angle DIC=\angle GIC+\angle DIG=\angle BIM+\angle EFA=\angle BIM+\angle MIA'=\angle BIA'\]as desired. Then, \[90 + 3\angle C/2 +\angle BID = \angle AIA'=\angle AIC+\angle CIA'=90+\angle B/2 + \angle CIA'.\] Case 2: $CF=CE$. This reduces to $BDIH$ cyclic where $H=AB\cap SI$. Let he incircle of $ABC$ touch $AB$ and $BC$ at $T$ and $J$. Thus \[\angle SIJ = \angle HIJ=\angle DIT = \angle APS=\angle AIH\]so $IS$ bisects $\angle AIJ$. Let $N$ be the other intersection of $(BDIH)$ and $(IA)$. Then, $BNS$ are collinear by Reim so $\angle AIN=\angle C$. Since $J$ is between $B$ and $A$, e have that $N$ is on minor arc $SJ$. Thus $\angle C = \angle AIN>\angle AIS = 45^{\circ}-\angle A/4$ which rearranges to the desired.