Let $ ABC$ be a triangle in which $ AB \equal{} AC$ and let $ I$ be its in-centre. Suppose $ BC \equal{} AB \plus{} AI$. Find $ \angle{BAC}$
1. Given that \( \triangle ABC \) is isosceles with \( AB = AC = b \) and \( BC = a + a = 2a \). Let \( I \) be the incenter of the triangle, and \( AI \) intersect \( BC \) at \( D \). We know \( BD = DC = a \) and \( AD = h \). Also, \( DI = r \) where \( r \) is the inradius.

2. By the Angle Bisector Theorem, we have:
   \[
   \frac{AI}{DI} = \frac{AB}{BD}
   \]
   Substituting the known values:
   \[
   \frac{AI}{r} = \frac{b}{a}
   \]
   Since \( r = \frac{S}{s} \) where \( S \) is the area of the triangle and \( s \) is the semi-perimeter, we have:
   \[
   r = \frac{ah}{a+b}
   \]
   Therefore:
   \[
   AI = \frac{bh}{a+b}
   \]

3. Given \( BC = AB + AI \), we substitute the known values:
   \[
   2a = b + \frac{bh}{a+b}
   \]
   Multiplying both sides by \( a + b \):
   \[
   2a(a + b) = b(a + b) + bh
   \]
   Simplifying:
   \[
   2a^2 + 2ab = ab + b^2 + bh
   \]
   Rearranging terms:
   \[
   2a^2 + ab = b^2 + bh
   \]
   Factoring out \( b \) on the right-hand side:
   \[
   2a^2 + ab = b(b + h)
   \]
   This is the equation \((*)\) given in the problem.

4. Next, consider points \( M \) and \( N \) such that \( M \in (BA \) produced and \( AM = AD \), and \( N \in (BC \) produced and \( CN = CA \). Since \( \triangle AMD \) and \( \triangle CAN \) are isosceles, we have:
   \[
   m(\widehat{AMD}) = \frac{m(\widehat{BAD})}{2}
   \]
   and
   \[
   m(\widehat{ACN}) = \frac{m(\widehat{ACB})}{2}
   \]

5. From the cyclic quadrilateral \( MADN \), we know that opposite angles sum to \( 180^\circ \). Therefore:
   \[
   \widehat{BAD} = \widehat{ACB}
   \]
   Since \( \triangle ABC \) is isosceles with \( AB = AC \), we have:
   \[
   \widehat{ACB} = 45^\circ
   \]

The final answer is \( \boxed{45^\circ} \).