Let $T$ be a point on a line segment $AB$ such that $T$ is closer to $B$ than to $A$. Show that for each point $C \ne T$ on the line through $T$ perpendicular to $AB$ there is exactly one point $D$ on the line segment $AC$ with $\angle CBD=\angle BAC$. Moreover, show that the line through $D$ perpendicular to $AC$ intersects the line $AB$ in a point $E$ which is independent of the position of $C$.
First $AC$ cuts the circle $(O)$ through $A,B$ and tangent to $CB$ at the desired point $D$ .
Let $F= DE\cap CT,G=DE\cap (ADB)$ it s clear that $ ADTF$ is cyclic then by Reim's we get $GB \parallel EF \implies BG\perp AB$ thus $O$ is midpiont of $FA$ . More $\angle DFT=\angle DAT =\angle DBC$ hence $DFBC$ is cyclic thus $\angle FBC=90^\circ$ which means that $B,F,O$ are colinear ,applying Menelaus's in the triangle $ AEG$ with $B-F-O$ we deduce $ \frac{BE}{BA}=\frac{EF}{FG}=\frac{ET}{TB}$ i.e. $\frac{BE}{ET}=\frac{BA}{TB}$ that s means that the point $E$ is fixed.
RH HAS