43. In a right triangle $ABC$, the bisector $CL$ is drawn. It is known that point $L$ is equidistant from the vertex of the right angle $B$ and the midpoint of the hypotenuse $AC$. Find the angle $BAC$.
$\square$ Answer: $30^{\circ}$.

Let $M$ be the midpoint of $AC$. Drop a perpendicular $LN$ from point $L$ to the hypotenuse $AC$. Since $L$ lies on the bisector of angle $BCA$, then $LB = LN$. By the condition, $LB = LM$. Therefore, points $M$ and $N$ coincide. Fig. 8. Then $LM$ is both a median and an altitude in triangle $ALC$, i.e., this triangle is isosceles. Therefore, $\angle LAC = \angle ACL = \angle BCL$. At the same time, $\angle LAC + \angle ACL + \angle BCL = 90^{\circ}$. Thus, $\angle BAC = \angle LAC = 30^{\circ}$.

(F. Nilov)