15. $96.5^{\circ}$
15. Extend $C A$ to $D$ so that $A D=A I$. Join $I B, I C$ and $I D$. Then we have $B C=A C+A I=A C+A D=C D$. It follows that $\triangle I B C$ and $\triangle I D C$ are congruent and so
$$
\angle B A C=2 \angle I A C=4 A D I=4 \angle I B C=2 \angle A B C .
$$

Let $\angle B A C=x$. Then $\angle A B C=\frac{x}{2}$ and $\angle A C B=\frac{x}{2}-13^{\circ}$. It follows that $x+\frac{x}{2}+\left(\frac{x}{2}-13^{\circ}\right)=180^{\circ}$, which gives $x=96.5^{\circ}$.
15. Let $I$ be the in-centre of $\triangle A B C$. If $B C=A C+A I$ and $\angle A B C-\angle A C B=13^{\circ}$, find $\angle B A C$.
(2 marks)
設 $I$ 為 $\triangle A B C$ 的內心。若 $B C=A C+A I$, 且 $\angle A B C-\angle A C B=13^{\circ}$, 求 $\angle B A C$.