Of course, we complex bash. Let $\omega$ be the unit circle and lines $BC$ and $DE$ be the tangents at $1$ and $-1$, respectively. Let $P$ and $Q$ be the tangents of $AB$ and $AC$ to $\omega$, respectively. Recall the so-called ``ice cream cone'' formula: if $z,w$ lie on the unit circle, then the intersection of their tangents is $\frac{2zw}{z + w}$. We apply ice cream cone four times to get \[a = \frac{2pq}{p + q} \qquad d = -\frac{2q}{q - 1} \qquad b = \frac{2p}{p + 1} \qquad c = \frac{2q}{q + 1}.\]Also, $m = \frac12d = -\frac q{q - 1}$. To verify $\angle AMD = \angle DBC$, we must show \[\frac{a - m}{d - m} \div \frac{d - b}{b - c} \in \mathbb R.\]So we compute \begin{align*} \frac{a - m}{d - m} \div \frac{d - b}{b - c} &= \frac{\frac{2pq}{p + q} + \frac q{q - 1}}{-\frac{2q}{q - 1} + \frac q{q - 1}} \cdot \frac{\frac{2p}{p + 1} - \frac{2q}{q + 1}}{-\frac{2q}{q - 1} - \frac{2p}{p + 1}} \\ &= \frac{2pq(q - 1) + q(p + q)}{q(p + q)} \cdot \frac{(q - 1)(p(q + 1) - q(p + 1))}{(q + 1)(q(p + 1) + p(q - 1))} \\ &= \frac{2pq - p + q}{p + q} \cdot \frac{(q - 1)(p - q)}{(q + 1)(2pq - p + q)} \\ &= \frac{(q - 1)(p - q)}{(p + q)(q + 1)}. \end{align*}This is its own conjugate since \[\overline{\left(\frac{(q - 1)(p - q)}{(p + q)(q + 1)}\right)} = \frac{(\frac1q - 1)(\frac1p - \frac1q)}{(\frac1p + \frac1q)(\frac1q + 1)} \cdot \frac{q \cdot pq}{pq \cdot q} = \frac{(1 - q)(q - p)}{(p + q)(1 + q)},\]therefore it is real.

Complex bash is for people who remember what an ice cream cone is Let $\angle A=2a$, $\angle B=2b$, $\angle C=2c$, let $\ell$ intersect $AB$ at $Y$ and extend $DI$ to intersect $BC$ at $X$. We make the stronger claim that $\triangle ADM \sim \triangle DXB$. Note that $I$ is the A-excenter of $\triangle AYD$. Using the bisections, we can find that \[\angle XDC=\frac{1}{2}(180-\angle ADY)=90-c.\]Since $\angle DCX=2c$, we get that $\angle DXC=90-c$, so $CDX$ is isosceles with $CD=CX$. Additionally, since $I$ is the incenter of $ABC$, $CI$ bisects $\angle C$ so $\angle ICD=c$. Therefore we have that $\angle CID=90$ and since $CDX$ is isosceles with $CD=CX$, this means that $I$ is the midpoint of $DX$. Additionally, note that \[\angle DXC=\angle XDC \implies \angle ADM=\angle DXB,\]so to prove our similarity, it suffices to show that \[\frac{AD}{DX}=\frac{DM}{BX}.\]Note that $DX=2*ID$ and $DM=\frac{ID}{2}$, so substituting this in gives us that this is equivalent to showing that \[\frac{AD}{ID}=\frac{ID}{BX}=\frac{IX}{BX},\]since $IX=BX$. However, note that $I$ is the incenter, so \[\angle IAD=a,\]\[\angle AID=180-\angle IAD-\angle ADM=180-a-(90+c)=b,\]and similarly, $\angle IBX=b$ and $\angle BIX=a$. This means that $\triangle AID \sim \triangle IBX$, so \[\frac{AD}{ID}=\frac{IX}{BX},\]as desired. Therefore $\triangle ADM \sim \triangle DXB$, and this means that \[\angle AMD=\angle DBX=\angle DBC, \]as desired. This finishes our proof.

If $DI \cap BC=K$ and $L$ is its reflection with respect to $B$, then $\triangle AID \cup \{M\} \sim \triangle DKL \cup \{B\}$ (since $IB$ is midline in $\triangle DKL$, $\angle IBK=\angle DLK=\angle AID$ and $\angle ADI=\angle DKL$ as $CD=CK$), which finishes the problem.

complex is easier but trig is more satisfying Construct $K$ such that $K \in ID$, $\angle AKM = C$, it is sufficient to prove that $\triangle AKM \sim \triangle DCB$. By SAS similarity, all we have to do is prove $\frac{KM}{AK} = \frac{BC}{CD}$. We can compute $CD = \frac{2r}{h} AC, AK = \frac{AD \sin 90 - \frac C2}{\sin C} = \frac{AC(h - 2r)}{2h \sin \frac C2}, KM = KD + DM = AK + \frac{r}{2 \cos \frac C2}$. We then just want to prove $1 + \frac{hr \sin \frac C2 }{AC(h - 2r) \cos \frac C2} = \frac{BC \cdot h}{2r \cdot AC}$, which resolves to $2r\cdot AC (h - 2r) \cos \frac C2 + 2hr^2 \sin \frac C2 = BC \cdot h(h -2r) \cos \frac C2$, replacing $h = AC \sin C$ gives the desired as $2r AC^2 \sin C \cos \frac C2 - 4r^2 AC \cos \frac C2 + 2 AC \cdot r^2 \sin \frac C2 \sin C = BC \cdot AC^2 \sin^2 C \cos \frac C2 - 2r \cdot BC \cdot AC \sin C \cos \frac C2$, dividing by $AC\cos \frac C2$ gives $2r AC \sin C - 4r^2 + 4r^2 \sin^2 \frac C2 = BC \cdot AC \sin^2 C - 2r \cdot BC \sin C$, which is just $r AC \sin \frac C2 - r^2 \cos \frac C2 = BC \cdot AC \sin^2 \frac C2 \cos \frac C2 - r BC \sin \frac C2$, rewriting as $rAC \sin \frac C2 - r^2 \cos \frac C2 = BC \cdot AC \frac{\sin^2 \frac C2}{\cos \frac C2} - r \cdot BC \sin \frac C2 - BC \cdot AC \frac{\sin^4 \frac C2 }{\cos \frac C2}$, or $(BC \sin \frac C2 - r \cos \frac C2)(AC \sin \frac C2 - r \cos \frac C2) = BC \cdot AC \sin^4 \frac C2$. Dividing by $\sin^2 \frac C2$ gives the desired as $(s - a)(s - b) = BC \cdot AC \sin^2 \frac C2$. We then prove that $(s - a) = 4R \cos \frac A2 \sin \frac B2 \sin \frac C2$ and the symmetrical variant, which is sufficient as we can reduce $BC = 2R \sin A = 4R \sin \frac A2 \cos \frac A2$, and the symmetrical variant, upon which the desired is trivial to prove. Note that we can write $2R^2 \sin A \sin B \sin C = \frac{abc}{4R} = A = rs = \frac Rr (\sin A + \sin B + \sin C) = Rr (2 \sin \frac{A + B}{2} \cos \frac{A - B}{2} + 2 \sin \frac{A + B}{2} \cos \frac{A + B}{2}) = 2Rr \cos \frac C2 (\cos \frac{A+ B}{2} + \cos \frac{A-B}{2}) = 4Rr \cos \frac C2 \cos \frac A2 \cos \frac B2$, then $r = 4R \sin \frac A2 \sin \frac B2 \sin \frac C2$, upon which the conclusion is obvious from considering the triangles formed by the incenter tangents and incenter.

here's the solution I found on the contest ignore the comment hehe

Just observe $\triangle AIB \sim \triangle ADI$ by angles, this motivates to put the midpoint of $BI$ and then just quickly angle chase everything. (If $N$ is the new midpoint, then note $ANIM$ is cyclic or that $\triangle ANM$ is similar to the above two triangles.)

iStud wrote: here's the solution I found on the contest ignore the comment hehe a bit remark for this solution. the main motivation is to see the midpoint condition and the angle-chasing needs will most likely pertain to finding common tangent for two circles. after some trial and errors, we suspect that if $F=BD\cap(AIB)$ then $(AFD)$ and $(ABIF)$ has a common tangent $DI$. then we can prove that $DFMQ$ is cyclic and tangent to $DE$, so it's left to simple angle-chasing only.