$ (O)$ is the circumcircle of the $ \triangle ABC,$ $ (M)$ is the circumcircle of $ B ICI_a,$ $ M \in (O),$ $ M$ is the midpoint of $ II_a.$ Regardles of the problem proposition, $ \angle AI_aB = \frac {\angle C}{2}.$ Reflect the circumcircle in the angle bisector $ AII_a$ into a circle $ (O'),$ $ O' \in AD,$ because $ AD, AO$ are isogonals WRT the angle $ \angle A.$ Also regardles of the problem proposition, let $ AL$ be a diameter of $ (O'),$ $ L \in AD,$ $ AL = 2R.$ By symmetry, $ (O'), (M)$ intersect at $ B' \in AC, C' \in AB,$ the reflections of $ B, C$ in $ AI.$ $ BDLC'$ is cyclic because of the right angles at $ C', D,$ $ \overline{AB} \cdot \overline {AC'} = \overline {AD} \cdot \overline {AL}$ and this is equal to power of $ A$ to $ (M),$ hence $ \overline{AD} \cdot \overline {AL} = AM^2 - MI^2.$ Since $ M \in (O'),$ $ ML \perp AM.$ Then $ I_aL^2 = IL^2 = MI^2 + ML^2 = MI^2 + AL^2 - AM^2 = AL^2 - \overline{AD} \cdot \overline {AL} = \overline{AL} \cdot \overline{DL}.$ It follows that the $ \triangle AI_aL \sim \triangle I_aDL$ with the common angle at $ L$ are similar by SAS and $ \angle DI_aL = \angle I_aAD = \frac {\angle B - \angle C}{2}.$ Let $ I_aL$ cut $ AB$ at $ K.$ $ I_aD$ bisects the angle $ \angle AI_aB\ \Longleftrightarrow$ $ \angle AI_aD = \frac {\angle C}{4}\ \Longleftrightarrow$ $ \angle AI_aK = \angle AI_aL = \angle AI_aD + \angle DI_aL = \frac {\angle B}{2} - \frac {\angle C}{4}\ \Longleftrightarrow$ ${ \angle AKI_a = 180^\circ - \left(\frac {\angle A}{2} + \frac {\angle B}{2} - \frac {\angle C}{4}\right) = 90^\circ + \frac {3}{4}\angle C}.$

Denote the interior angles of $\Delta ABC$ simply as $\angle A, \angle B, \angle C$. Let $O$ be the circumcenter, $I$ the incenter of $ABC$, the incircle touches $BC, CA$ at $E, F$, respectively; $AI$ meets the circumcircle again at $P$, $AO$ meets the circumcircle again at $Q$. $QI$ meets the circumcircle again at $H$. Let the A-excircle with center $I_{a}$ touches $AB, BC, CA$ at $G, R, S$, respectively. Notice that $\angle QAC = \angle BAD = 90^{\circ} - \angle B$, so $AI$ is also the angle bisector of $\angle DAQ$. First we are going to show that $\angle DI_{a}A = \angle RI_{a}Q$ It is well known that $H, E, P$ are collinear (Let $PO$ meets the circumcirle again at $T$. Notice that $PB = PI = PC$. As $angle EIP = \angle IAQ$ and $IE/AI = IF/AI = sin(A/2)$ and $IP/AQ = PB/PT = sin(A/2), \Delta IEP\sim \Delta AIQ$, thus $\angle IQA = \angle IPE$, then $H, E, P$ are collinear). By sinus law in $\Delta AI_{a}C$ we have $AI_{a}/cos(C/2) = AC/sin(B/2) = AQ.sinB/sin(B/2)$ and in $\Delta AIB$: $AI/sin(B/2) = AB/cos(C/2) = AD/sinB.cos(C/2)$. Thus $AI/AQ = AD/AI_{a},\Delta AI_{a}Q\sim \Delta AID$, this means first $\angle AI_{a}Q = \angle ADI$. Second, if $AD$ meets $PH$ at $U$ then easy to see that $HAIU$ is cyclic and $\angle AUI = 90^{\circ}$, then $UIED$ is a rectangle. Then $\angle AI_{a}Q = \angle ADI = \angle UEI$, it follows that $\angle RI_{a}Q = \angle IPE$ and $AD/AI_{a} = AI/AQ = AU/AP, UP\parallel DI_{a}$. Therefore $\angle DI_{a}A = \angle EPI = \angle RI_{a}Q$. Now easy to see that $AL = 2R = AQ\iff L$ is the reflection of $Q$ over $AI_{a}$ (then of course $L, P, Q$ collinear) $\iff \angle QI_{a}S = \angle KI_{a}G = 3/4 C\iff I_{a}Q$ is the angle bisector of $\angle RI_{a}C \iff \angle DI_{a}A = C/4\iff DI_{a}$ is the angle bisector of $\angle BI_{a}A (= C/2)$.

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It is not hard,if we know this problem: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1178972&sid=33b04ae6e251ae0fc66cc926e418dda0#p1178972