Let $R=BP\cap CQ$, then $\angle PRQ=60^\circ$, $\angle PIQ = \angle PHQ = 120^\circ$. Hence $P,Q,R,I,H$ are concyclic. Also, $\triangle RBC$ is an equilateral triangle and $M$ is the midpoint of $OR$. Note that $MN\parallel RI$ and $\angle AOQ = \frac{1}{2}\angle AOC = \angle ABC = \angle ROP$ implies that $OI$ is the angle bisector of $\angle AOM$. Hence, \begin{align*} \measuredangle(HI,DA) &= \measuredangle HIR + \measuredangle MDA = \measuredangle HPR + \measuredangle IOA \\ &= \measuredangle QOB + \measuredangle QOA + \measuredangle IOQ = \measuredangle QOB + \measuredangle COQ - \measuredangle QOI \\ &= \measuredangle COB - \measuredangle QOI = 120^\circ - 60^\circ /2 = 90^\circ \end{align*}(Counterclockwise directed angle $\text{mod} 180^\circ$)

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This can be reduced the following nicer-looking statement by angle chasing: Let $ABC$ be an acute triangle with $AB<AC$ and $\angle BAC=60^\circ$. Let $I$ be the incenter of $\triangle ABC$, let $D$ be the foot of the altitude from $A$ to $\overline{BC}$, and let $M$ be the midpoint of $\overline{AI}$. Prove that $\angle ADM=\frac{\angle ABC}{2}-30^\circ$. Here's a (edit: way too complicated) solution to the modified problem: Let the incircle of $\triangle ABC$ touch sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ at $K$, $E$, and $F$, respectively. Since $\angle EMI=\angle FMI=2\angle EAI=60^\circ$, we see that $EIM$ and $FIM$ are equilateral. Thus, $IE=IF=IM$, so $M$ is on the incircle of $\triangle ABC$. Let $\overline{AD}$ intersect the circumcircle of $AEIF$ again at $P$. Then, notice that $\angle API=\angle ADK=\angle DKI=90^\circ$, so $DKIP$ is a rectangle. Thus, $MP=MI=IK=DP$, so $\triangle PDM$ is isosceles. We have \[2\angle ADM=\angle APM=\angle PAM=\angle BAI-\angle BAD=30^\circ-(90^\circ-\angle ABC)=\angle ABC-60^\circ.\]Therefore, $\angle ADM=\frac{\angle ABC}{2}-30^\circ$.

CyclicISLscelesTrapezoid wrote: Let $ABC$ be an acute triangle with $AB<AC$ and $\angle BAC=60^\circ$. Let $I$ be the incenter of $\triangle ABC$, let $D$ be the foot of the altitude from $A$ to $\overline{BC}$, and let $M$ be the midpoint of $\overline{AI}$. Prove that $\angle ADM=\frac{\angle ABC}{2}-30^\circ$. Another finish is as follows: $AI \cap BC=K, AI \cap (ABC)=N, AA'$ is diameter. If the inner angle bisector of $DAK$ intersect $BC$ at $X$ it is enough to prove that $\angle DAK=2\angle ADM \iff DM \text{ bisects } AX \iff IX \parallel MD \iff \frac{AD}{AK}=\frac{MI}{IK} \iff \frac{2AD}{AK}=\frac{AI}{IK} \iff \frac{2AN}{AA'}=\frac{AB+AC}{BC} \iff R=\frac{AN\cdot BC}{AB+AC} \overset{Ptolemy} \iff R=NB \iff \angle A=60^\circ$.

Let BP and CQ meet at S. Let HI meet MD at K. Claim1 : SPIHQ is cyclic. Proof : ∠PSQ = 180 - ∠BOC = 60 and ∠QHP = 180 - ∠QOP = 180 - ∠COB/2 = 120 so PSQH is cyclic. ∠QIP = 90 + ∠QOP/2 = 120 so PSQI is cyclic. Claim2 : MN || SI. Proof : ∠OBS = 90 and ∠BSO = 30 so SO = 2BO and we know BO = MO so MO = MS. MO = MS and NI = NO so MN || SI. Claim3 : ∠ADK + ∠DKH = 90. Proof : ∠ADK = 30 - ∠ABC and ∠DKH = ∠KIS = ∠HQS = ∠HQA + 2∠ABC so ∠ADK + ∠DKH = 30 + ∠AQH + ∠ABC so we need to show ∠AQH + ∠ABC = 60 or in fact ∠AQH = ∠ACB. ∠AQH = ∠HOP = ∠AOB/2 = ∠ACB. we're Done.

Quite a problem. We solve the modified version. Let the incircle touch $\overline{AB}$ and $\overline{AC}$ at $Y$ and $X$, respectively. Furthermore set $K = \overline{AD} \cap (AXY)$. Notice that $MY=MI=MK$ because $\angle A = 60^\circ$, and furthermore $\overline{AK} \perp \overline{KI}$, so thus $KIED$ is a rectangle and $KD=KM$. As a result, $\angle ADM = \frac 12 \angle KAM = \frac 12 \angle ABC - 30^\circ$, as needed.