1. Take a point $A_{1}$ on the line $B A$ such that $\left|A_{1} B\right|=\left|A_{1} C\right|$. The points $A_{1}, A, D$ and $C$ lie on the same circle $\left(\widehat{D A_{1} C}=90^{\circ}-\widehat{A B C}=\widehat{D A C}\right)$. Therefore, $\widehat{A_{1} A C}=\widehat{A_{1} D C}=90^{\circ}$, which means $\widehat{B A C}=90^{\circ}$.
1. In triangle $ABC$, the median $AD$ is drawn. $\widehat{D A C} + \widehat{A B C} = 90^{\circ}$. Find $\widehat{B A C}$, given that $|A B| = |A C|$.