Sharygin 2013 8.5 wrote: The altitude $AA'$, the median $BB'$, and the angle bisector $CC'$ of a triangle $ABC$ are concurrent at point $K$. Given that $A'K = B'K$, prove that $C'K = A'K$. It's easy to notice that $BB'\perp AC$ by just congruence of the triangles $B'KC$ and $A'KC$. Again notice the congruence of the triangles $ABB'$ and $CBB'$. So, $AB=BC$ also by the congruence of the triangles $ACC'$ and $BCC'$ we get that $AC=BC$. So, $AB=BC=CA$, now the conclusion follows trivially. BTW Here is an interesting fact about this Configuration. If in a Triangle, Altitude $AA'$, Median $BB'$ and angle bisector $CC'$ are concurrent then $$\boxed{a^2(a-b)=(c^2-b^2)(a+b)}$$ @below you have to adjust your diagram, if you see very closely then the $C-$ bisector is not going through $K$.

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