8.3. On the side $AC$ of triangle $ABC$, a point $M$ is taken. It turns out that $AM = BM + MC$ and $\angle BMA = \angle MBC + \angle BAC$. Find $\angle BMA$.
Answer: $60^{\circ}$. Solution. First, we show that triangle $A B C$ is isosceles. Indeed, this follows from the condition $\angle B M A=\angle M B C+\angle B A C$ and the property of the exterior angle: $\angle B M A=\angle M B C+\angle B C A$. From these two equalities, we have $\angle B C A=\angle B A C$, and thus, triangle $A B C$ is isosceles. Next, we take a point $K$ on side $A C$ that is symmetric to point $M$ with respect to point $O$ - the midpoint of $A C$. Then $A K=M C$ and therefore from the relation $A M=B M+M C$ it follows that point $M$ lies between $C$ and $O$ and, hence, this relation gives $K M=B M$. Since $\triangle A B C$ is isosceles, point $O$ coincides with the projection of point $B$ onto the base $A C$, and therefore $B M=B K$. Thus, in triangle $K B M$ all sides are equal, meaning it is equilateral and all its angles are $60^{\circ}$.