APMO 2021

Let $L$ be the intersection of the bisectors of $\angle ABC$ and $\angle BCD$. Let $N$ be the $E$-excenter of $△BCE$. Let $\angle BAC=\angle BDC=\alpha$, $\angle DBC=\beta$ and $\angle ACB=\gamma$. We have the following: $$\begin{array}{r}\angle CBL=\frac{1}{2}\angle ABC=90^{\circ }-\frac{1}{2}\alpha -\frac{1}{2}\gamma \text{ and }\angle BCL=90^{\circ }-\frac{1}{2}\alpha -\frac{1}{2}\beta \\ \angle CBN=90^{\circ }-\frac{1}{2}\beta \text{ and }\angle BCN=90^{\circ }-\frac{1}{2}\gamma \\ \angle MBL=\angle MBC+\angle CBL=90^{\circ }-\frac{1}{2}\gamma \text{ and }\angle MCL=90^{\circ }-\frac{1}{2}\beta \\ \angle LCN=\angle LBN=180^{\circ }-\frac{1}{2}(\alpha +\beta +\gamma )\end{array}$$ Applying the sine rule to $△MBL$ and $△MCL$ we obtain $$\frac{MB}{ML}=\frac{MC}{ML}=\frac{\sin \angle BLM}{\sin \angle MBL}=\frac{\sin \angle CLM}{\sin \angle MCL}$$ It follows that $$\begin{array}{c}\frac{\sin \angle BLM}{\sin \angle CLM}=\frac{\sin \angle MBL}{\sin \angle MCL}=\frac{\cos (\gamma /2)}{\cos (\beta /2)}\end{array}\text{\hspace{1em}(1)}$$ Now $$\frac{\sin \angle BLM}{\sin \angle MLC}\cdot \frac{\sin \angle LCN}{\sin \angle NCB}\cdot \frac{\sin \angle NBC}{\sin \angle NBL}=\frac{\cos (\gamma /2)}{\cos (\beta /2)}\cdot \frac{\sin \left(90^{\circ }-\frac{1}{2}\beta \right)}{\sin \left(90^{\circ }-\frac{1}{2}\gamma \right)}=1.$$ Hence $LM$, $BN$, $CN$ are concurrent and therefore $L$, $M$, $N$ are collinear.

We proceed similarly as above until the equation (1). We use the following lemma. Lemma: If $\pi >\alpha ,\beta ,\gamma ,\delta >0$, $\alpha +\beta =\gamma +\delta <\pi$, and $\frac{\sin \alpha }{\sin \beta }=\frac{\sin \gamma }{\sin \delta }$, then $\alpha =\gamma$ and $\beta =\delta$. Proof of Lemma: Let $\theta =\alpha +\beta =\gamma +\delta$. Then $\frac{\sin (\theta -\beta )}{\sin \beta }=\frac{\sin (\theta -\delta )}{\sin \delta }$. $$\begin{array}{c}⟺\sin (\theta -\beta )\sin \delta =\sin (\theta -\delta )\sin \beta \\ ⟺(\sin \theta \cos \beta -\sin \beta \cos \theta )\sin \delta =(\sin \theta \cos \delta -\sin \delta \cos \theta )\sin \beta \\ ⟺\sin \theta \cos \beta \sin \delta =\sin \theta \cos \delta \sin \beta \\ ⟺\sin \theta \sin (\beta -\delta )=0\end{array}$$ Since $0<\theta <\pi$, then $\sin \theta \ne 0$. Therefore, $\sin (\beta -\delta )=0$, and we must have $\beta =\delta$. Applying the sine rule to $△NBL$ and $△NCL$ we obtain $$\begin{array}{rl}& \frac{NB}{NL}=\frac{\sin \angle BLN}{\sin \angle LBN}\\ & \frac{NC}{NL}=\frac{\sin \angle CLN}{\sin \angle LCN}\end{array}$$ Since $\angle LBN=\angle LCN$, it follows that $$\frac{\sin \angle BLN}{\sin \angle CLN}=\frac{NB}{NC}=\frac{\sin \angle BCN}{\sin \angle CBN}=\frac{\cos (\gamma /2)}{\cos (\beta /2)}=\frac{\sin \angle BLM}{\sin \angle CLM}$$ By the lemma, it is concluded that $\angle BLM=\angle BLN$ and $\angle CLM=\angle CLN$. Therefore, $L$, $M$, $N$ are collinear.

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Alternative solution.

Denote by $N$ the excenter of triangle $BCE$ opposite $E$. Since $BL$ bisects $\angle ABC$, we have $\angle CBL=\frac{\angle ABC}{2}$. Since $M$ is the midpoint of arc $BC$, we have $\angle MBC=\frac{1}{2}(\angle MBC+\angle MCB)$ It follows by angle chasing that $$\begin{array}{rl}\angle MBL& =\angle MBC+\angle CBL=\frac{1}{2}(\angle MBC+\angle MCB+\angle ABC)\\ & =\frac{1}{2}(\angle MBA+\angle MCB)=90^{\circ }-\frac{\angle BCE}{2}=\angle BCN\end{array}$$ Denote by $X$ and $Y$ the second intersections of lines $BM$ and $CM$ with the circumcircle of $BCL$, respectively. Since $\angle MBC=\angle MCB$, we have $BC\parallel XY$. It suffices to show that $BN\parallel XL$ and $CN\parallel YL$. Indeed, from this it follows that $△BCN\sim △XYL$, and therefore a homothety with center $M$ that maps $B$ to $X$ and $C$ to $Y$ also maps $N$ to $L$, implying that $N$ lies on the line $LM$. By symmetry, it suffices to show that $CN\parallel YL$, which is equivalent to showing that $\angle BCN=\angle XYL$. But we have $\angle BCN=\angle MBL=\angle XBL=\angle XYL$, completing the proof.