(i) $ABC$ is a triangle with a right angle at $A$, and $P$ is a point on the hypotenuse $BC$. 
The line $AP$ produced beyond $P$ meets the line through $B$ which is perpendicular to $BC$ at $U$. 
Prove that $BU = BA$ if, and only if, $CP = CA$.
(ii) $A$ is a point on the semicircle $CB$, and points $X$ and $Y$ are on the line segment $BC$. 
The line $AX$, produced beyond $X$, meets the line through $B$ which is perpendicular to $BC$ at $U$. 
Also the line $AY$, produced beyond $Y$, meets the line through $C$ which is perpendicular to $BC$ at $V$. 
Given that $BY = BA$ and $CX = CA$, determine the angle $\angle VAU$.
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   - Since $BY = BA$, $\triangle BYA$ is an isosceles right triangle with $\angle BYA = 90^\circ$.
   - Since $CX = CA$, $\triangle CXA$ is an isosceles right triangle with $\angle CXA = 90^\circ$.
   - Since $A$ is on the semicircle $CB$, $\angle CAB = 90^\circ$.
   - Therefore, $\angle BAY = \angle BAX = 45^\circ$.
   - Since $U$ is the intersection of the line through $B$ perpendicular to $BC$ and $AX$ produced, $\angle BAU = 45^\circ$.
   - Similarly, since $V$ is the intersection of the line through $C$ perpendicular to $BC$ and $AY$ produced, $\angle CAV = 45^\circ$.
   - Therefore, $\angle VAU = \angle BAU + \angle CAV = 45^\circ + 45^\circ = 90^\circ$.

$\boxed{90^\circ}$