Question 167
Given a triangle $ABC$, let $D$ be the point of the ray $BA$ such that $BD = BA + AC$. If $K$ and $M$ are points on the sides $BA$ and $BC$, respectively, such that the triangles $BDM$ and $BCK$ have the same areas, prove that $\angle BKM = \frac{1}{2} \angle BAC$.