, 2021

Let $L$ be the intersection of the bisectors of $\angle ABC$ and $\angle BCD$. Let $N$ be the $E$-excenter of $\triangle BCE$. 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 $\triangle MBL$ and $\triangle MCL$ 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
$$\begin{equation*} \frac{\sin \angle BLM}{\sin \angle CLM}=\frac{\sin \angle MBL}{\sin \angle MCL}=\frac{\cos (\gamma / 2)}{\cos (\beta / 2)} \tag{1} \end{equation*}$$
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 $\frac{\sin (\theta-\beta)}{\sin \beta}=\frac{\sin (\theta-\delta)}{\sin \delta}$.
$$\begin{gathered} \Longleftrightarrow \sin (\theta-\beta) \sin \delta=\sin (\theta-\delta) \sin \beta \\ \Longleftrightarrow (\sin \theta \cos \beta-\sin \beta \cos \theta) \sin \delta=(\sin \theta \cos \delta-\sin \delta \cos \theta) \sin \beta \\ \Longleftrightarrow \sin \theta \cos \beta \sin \delta=\sin \theta \cos \delta \sin \beta \\ \Longleftrightarrow \sin \theta \sin (\beta-\delta)=0 \end{gathered}$$
Since $0<\theta<\pi$, then $\sin \theta \neq 0$. Therefore, $\sin (\beta-\delta)=0$, and we must have $\beta=\delta$.
Applying the sine rule to $\triangle NBL$ and $\triangle NCL$ we obtain
$$\begin{aligned} & \frac{NB}{NL}=\frac{\sin \angle BLN}{\sin \angle LBN} \\ & \frac{NC}{NL}=\frac{\sin \angle CLN}{\sin \angle LCN} \end{aligned}$$
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.

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
$$\begin{aligned} \angle MBL & =\angle MBC+\angle CBL=\frac{1}{2}(\angle MB C+\angle MC B+\angle ABC) \\ & =\frac{1}{2}(\angle MBA+\angle MCB)=90^\circ-\frac{\angle BCE}{2}=\angle BCN \end{aligned}$$
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 $\triangle BCN \sim \triangle XYL$, 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.