Angles and Midpoints

[Thanic Nur Samin]

Let $B$ and $C$ be two points and $M$ be their midpoint. Let $X$ be also a point on the same plane. Define $T$ to be a point so that $\angle XMT=90^{\circ}$ and $XB=XT$ which lies inside the circumcircle of $BMX$. Prove that, $\angle TXB=2\angle MTC$.

[Thanic Nur Samin]

Let $N$ denote the midpoint of $BT$. So, $\angle TNX=90^{\circ}, \angle TMX=90^{\circ}$. So $TNMX$ is cyclic. Now, $\dfrac{1}{2}\angle BXT=\angle TXN=\angle TMN$. Since $M$ and $N$ are midpoints of $BC$ and $BT$ respictively, $MN\parallel CT$, so we get $\angle TMN=\angle MTC$, therefore $\angle BXN=2\angle CTM$.