Question: The points $M$ and $N$ are chosen on the angle bisector $AL$ of a $\Delta ABC$ such that $\angle ABM=\angle ACN=23^0$. $X$ is a point inside the triangle such that $BX=CX$ and $\angle BXC = 2\angle BML$. Find $\angle MXN$.
Solution: Let $AL$ meet the circumcircle of $\Delta ABC$ at $D\neq A$. Join the points $A,D$; $C,D$; $X,D$.
Let us assume that $\angle BML=\theta.$ This implies that $\angle BXC=2\theta$.
Now since $\angle ABM=23^0$ and $\angle BML=\theta\implies \angle BAL=\theta-23^0.$ This in turn implies that $\angle LAC=\angle BAL=\theta-23^0$.
Again since $\angle ACN=23^0$ and $\angle CAN=\theta-23^0\implies \angle CNL=\theta.$
Also observe that $\angle DBC=\angle DCB=\theta-23^0\implies DB=DC.$ Using this we can conclude that $\Delta XBD\cong \Delta XCD$. This implies that $\angle BXD=\angle CXD=\theta$.
Let us assume that $BC$ intersects $XD$ at $J$. Now since $XJ\perp BC\implies \angle XBJ=\angle XCJ=90^0-\theta.$ Thus, we can conclude that $\angle XBD=\angle XCD=(90^0-\theta)+(\theta-23^0)=67^0.$
Now since $\angle DNC=\angle DXC=\theta$, implies that the points $D,N,X,C$ are concyclic. Again since $\angle BMD=\angle BXD=\theta$, implies that $B,M,X,D$ are concyclic.
Thus, $\angle DBX=\angle DMX=67^0$ and $\angle XCD=\angle MNX=67^0$.
Thus, we can conclude that $\angle MXN=46^0$.
Is there a better and shorter solution than this one?