Answer:
$$
53
$$

Solution:
Since $B E \perp A C, \angle B A E=90^{\circ}-\angle A B E=74^{\circ}$. Now, $n^{\circ}=180-\angle B X A=\angle E B A+\angle B A D=$ $16^{\circ}+\frac{74^{\circ}}{2}=53^{\circ}$.
1. Let $n$ be the answer to this problem. In acute triangle $A B C$, point $D$ is located on side $B C$ so that $\angle B A D=\angle D A C$ and point $E$ is located on $A C$ so that $B E \perp A C$. Segments $B E$ and $A D$ intersect at $X$ such that $\angle B X D=n^{\circ}$. Given that $\angle X B A=16^{\circ}$, find the measure of $\angle B C A$.
Proposed by: Joseph Heerens