I have the following setup.
A triangle $ABC$ has incentre $I$ and $AB>AC$. $E$ is a point on the circumcircle of $ABC$ such that $\angle IEA=90^\circ$. $U$ is a point on $AC$ such that $IU$ is perpendicular to $AC$, and $W$ is a point on $AB$ such that $IW$ is perpendicular to $AB$.
How do I prove that $\triangle EUC$ is similar to $\triangle EWB$?
Clearly we have for triangles $EUC$ and $EWB$, $$\angle ECU = \angle EBW$$ subtended by chord $AE$ on the circumcircle $(ABC)$.
Also $E,U,W$ lie on the circle with diameter $AI$. So $$\angle AUE= \angle AWE$$ $$\Rightarrow \angle EUC= \angle EWB$$
Thus $\triangle EUC \sim \triangle EWB$ by angle-angle similarity criterion.